{"id":121198,"date":"2022-05-13T16:49:19","date_gmt":"2022-05-13T11:19:19","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=121198"},"modified":"2022-05-13T16:49:19","modified_gmt":"2022-05-13T11:19:19","slug":"chapter-3-trigonometric-functions-questions-and-answers-ncert-solutions-for-class-11-maths","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-3-trigonometric-functions-questions-and-answers-ncert-solutions-for-class-11-maths","title":{"rendered":"Chapter 3 &#8211; Trigonometric Functions Questions and Answers: NCERT Solutions for Class 11 Maths"},"content":{"rendered":"<h2>Exercise 3.1<\/h2>\n<h2>1. Find the radian measures corresponding to the following degree measures:<br \/>\n(i) 25o25o<\/h2>\n<h3>Ans: We know that 180o=\u03c0180o=\u03c0 radian<br \/>\nTherefore 1\u2218=\u03c01801\u2218=\u03c0180 radian<br \/>\nhence,<br \/>\n25o=\u03c0180\u00d72525o=\u03c0180\u00d725 radian<br \/>\n=5\u03c036=5\u03c036radian<\/h3>\n<h2>(ii) -47o30\u2032-47o30\u2032<\/h2>\n<h3>Ans: Here we have,<br \/>\n-47o30\u2032=-4712o-47o30\u2032=-4712o<br \/>\n=-952=-952 degree<br \/>\nSince we know that, 180o=\u03c0180o=\u03c0 radian<br \/>\nTherefore 1o=\u03c01801o=\u03c0180 radian<br \/>\nHence,<br \/>\n-952-952 degree=\u03c0180\u00d7(-952)=\u03c0180\u00d7(-952) radian<br \/>\n=(-1936\u00d72)\u03c0=(-1936\u00d72)\u03c0 radian<br \/>\n=-1972\u03c0=-1972\u03c0radian<br \/>\nTherefore,<br \/>\n-47o30\u2032=-1972\u03c0-47o30\u2032=-1972\u03c0 radian<\/h3>\n<h2>(iii) 240o240o<\/h2>\n<h3>Ans: We know that,<br \/>\n180o=\u03c0180o=\u03c0 radian<br \/>\nTherefore 1o=\u03c01801o=\u03c0180 radian<br \/>\nHence,<br \/>\n240o=\u03c0180\u00d7240240o=\u03c0180\u00d7240 radian<br \/>\n=43\u03c0=43\u03c0radian<\/h3>\n<h2>(iv) 520o520o<\/h2>\n<h3>Ans: We know that,<br \/>\n180o=\u03c0180o=\u03c0 radian<br \/>\nTherefore 1o=\u03c01801o=\u03c0180 radian<br \/>\nHence,<br \/>\n520o=\u03c0180\u00d7520520o=\u03c0180\u00d7520 radian<br \/>\n=26\u03c09=26\u03c09radian<\/h3>\n<h2>2. Find the degree measures corresponding to the following radian measures<br \/>\n(Use\u03c0=227\u03c0=227 )<br \/>\n(i) 11161116<\/h2>\n<h3>Ans: We know that,<br \/>\n\u03c0\u03c0 radian=180o=180o<br \/>\nTherefore 1 radian =180\u03c0o1 radian =180\u03c0o<br \/>\nHence,<br \/>\n11161116 radian=180\u03c0\u00d71116=180\u03c0\u00d71116 degree<br \/>\n=45\u00d711\u03c0\u00d74=45\u00d711\u03c0\u00d74degree<br \/>\n=45\u00d711\u00d7722\u00d74=45\u00d711\u00d7722\u00d74<br \/>\ndegree<br \/>\n=3158=3158degree<br \/>\nFurther computing,<br \/>\n11161116 radian=3938=3938 degree<br \/>\n=39o+3\u00d7608=39o+3\u00d7608 minutes<br \/>\nSince 1o=60\u20321o=60\u2032<br \/>\n11161116 radian =39o+22\u2032+12=39o+22\u2032+12minutes<br \/>\nSince 1\u2032=60&#8221;1\u2032=60&#8221;<br \/>\n11161116 radian=39o22\u203230&#8221;=39o22\u203230&#8221;<\/h3>\n<h2>(ii) -4-4<\/h2>\n<h3>Ans: We know that,<br \/>\n\u03c0\u03c0 radian=180o=180o<br \/>\nTherefore 1 radian =180\u03c0o1 radian =180\u03c0o<br \/>\nHence,<br \/>\n-4-4 radian=180\u03c0\u00d7(-4)=180\u03c0\u00d7(-4) degree<br \/>\n=180\u00d77(-4)22=180\u00d77(-4)22degree<br \/>\n=-252011=-252011 degree<br \/>\n=-229111=-229111degree<br \/>\nSince 1o=60\u20321o=60\u2032<br \/>\nWe have,<br \/>\n-4-4 radian=-229o+1\u00d76011=-229o+1\u00d76011 minutes<br \/>\n=-229o+5\u2032+511=-229o+5\u2032+511 minutes<br \/>\nSince 1\u2032=60\u2032\u20321\u2032=60\u2032\u2032<br \/>\n-4-4 radian=-229o5\u203227&#8221;=-229o5\u203227&#8221;<\/h3>\n<h2>(iii) 5\u03c035\u03c03<\/h2>\n<h3>Ans: We know that,<br \/>\n\u03c0\u03c0 radian=180o=180o<br \/>\nTherefore 1 radian =180\u03c0o1 radian =180\u03c0o<br \/>\nHence,<br \/>\n5\u03c035\u03c03 radian=180\u03c0\u00d75\u03c03=180\u03c0\u00d75\u03c03 degree<br \/>\n=300o=300o<\/h3>\n<h2>(iv)7\u03c067\u03c06<\/h2>\n<h3>Ans: We know that,<br \/>\n\u03c0\u03c0 radian=180o=180o<br \/>\nTherefore 1 radian =180\u03c0o1 radian =180\u03c0o<br \/>\nHence,<br \/>\n7\u03c067\u03c06 radian=180\u03c0\u00d77\u03c06=180\u03c0\u00d77\u03c06<br \/>\n=210o=210o<\/h3>\n<h2>3. A wheel makes 360360 revolutions in one minute. Through how many radians does it turn in one second?<\/h2>\n<h3>Ans: Number of revolutions the wheel makes in 11 minute=360=360<br \/>\nNumber of revolutions the wheel make in 11 second=36060=36060<br \/>\n=6=6<br \/>\nIn one complete revolution, the wheel turns an angle of<br \/>\n2\u03c02\u03c0<br \/>\nradian.<br \/>\nHence, it will turn an angle of 6\u00d72\u03c0=12\u03c06\u00d72\u03c0=12\u03c0 radian, in 66 complete revolutions.<br \/>\nTherefore, the wheel turns an angle of 12\u03c012\u03c0 radian in one second.<\/h3>\n<h2>4. Find the degree measure of the angle subtended at the centre of a circle of radius 100100cm by an arc of length 2222 cm.<br \/>\n(Use\u03c0=227\u03c0=227 )<\/h2>\n<h3>Ans: We know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr \u2026\u2026(1)<br \/>\nTherefore,<br \/>\nSubstituting r=100cmr=100cm ,<br \/>\nl=22cml=22cm<br \/>\nin the formula (1) , we have,<br \/>\n\u03b8=22100\u03b8=22100 radian<br \/>\nSince 1 radian=180\u03c01 radian=180\u03c0<br \/>\nTherefore,<br \/>\n\u03b8=180\u03c0\u00d722100\u03b8=180\u03c0\u00d722100 degree<br \/>\n=180\u00d77\u00d72222\u00d7100=180\u00d77\u00d72222\u00d7100degree<br \/>\n=635=635 degree<br \/>\n=1235=1235degree<br \/>\nSince 1o=60\u20321o=60\u2032 , we have,<br \/>\n\u03b8=12o36\u2032\u03b8=12o36\u2032<br \/>\nHence , the required angle is 12o36\u203212o36\u2032.<\/h3>\n<h2>5. In a circle of diameter 4040 cm, the length of a chord is 2020 cm. Find the length of minor arc of the chord.<\/h2>\n<h3>Ans: Given that, diameter of the circle=40=40 cm<br \/>\nHence Radius (r)(r) of the circle=402cm=402cm<br \/>\n=20cm=20cm<br \/>\nLet ABAB be a chord of the circle whose length is 2020 cm.<\/h3>\n<h3>In \u0394OAB,\u0394OAB,<br \/>\nOA=OBOA=OB<br \/>\n== Radius of the circle<br \/>\n=20cm=20cm<br \/>\nNow also, AB=20cmAB=20cm<br \/>\nTherefore, \u0394OAB\u0394OAB is an equilateral triangle.<br \/>\n\u2234\u03b8=60o\u2234\u03b8=60o<br \/>\n=\u03c03=\u03c03 radian<br \/>\nWe know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr \u2026\u2026(1)<br \/>\nSubstituting \u03b8=\u03c03\u03b8=\u03c03 in the formula (1),<br \/>\n\u03c03=arc AB20\u03c03=arc AB20<br \/>\narc AB=20\u03c03cmarc AB=20\u03c03cm<br \/>\nTherefore, the length of the minor arc of the chord is 20\u03c03cm20\u03c03cm .<\/h3>\n<h2>6. If in two circles, arcs of the same length subtend angles 60o60o and 75o75o at the centre, find the ratio of their radii.<\/h2>\n<h3>Ans: Let the radii of the two circles be r1r1 and r2r2 . Let an arc of length l1l1 subtends an angle of 60o60o at the centre of the circle of radius r1r1 , whereas let an arc of length l2l2 subtends an angle of 75o75o at the centre of the circle of radius r2r2 .<br \/>\nNow, we have,<br \/>\n60o=\u03c0360o=\u03c03 radian and<br \/>\n75\u2218=5\u03c01275\u2218=5\u03c012 radian<br \/>\nWe know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr<br \/>\nl=r\u03b8l=r\u03b8<br \/>\nHence we obtain,<br \/>\nl=r1\u03c03l=r1\u03c03<br \/>\nand l=r25\u03c012l=r25\u03c012<br \/>\naccording to the l1=l2l1=l2<br \/>\nthus we have,<br \/>\nr1\u03c03\u03c0=r25\u03c012r1\u03c03\u03c0=r25\u03c012<br \/>\nr1=r254r1=r254<br \/>\nr1r2=54r1r2=54<br \/>\nHence , the ratio of the radii is 5:45:4 .<\/h3>\n<h2>7. Find the angle in radian through which a pendulum swings if its length is 7575 cm and the tip describes an arc of length.<br \/>\n(i) 1010 cm<\/h2>\n<h3>Ans: We know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr<br \/>\nGiven that r=75cmr=75cm<br \/>\nAnd here, l=10cml=10cm<br \/>\nHence substituting the values in the formula,<br \/>\n\u03b8=1075\u03b8=1075 radian<br \/>\n=215=215radian<\/h3>\n<h2>(ii) 1515 cm<\/h2>\n<h3>Ans: We know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr<br \/>\nGiven that r=75cmr=75cm<br \/>\nAnd here, l=15cml=15cm<br \/>\nHence substituting the values in the formula,<br \/>\n\u03b8=1575\u03b8=1575 radian<br \/>\n=15=15radian<\/h3>\n<h2>(iii) 2121 cm<\/h2>\n<h3>Ans: We know that,<br \/>\nin a circle of radius rr unit, if an angle \u03b8\u03b8 radian at the centre is subtended by an arc of length ll unit then<br \/>\n\u03b8=lr\u03b8=lr<br \/>\nAnd here, l=21cml=21cm<br \/>\nHence substituting the values in the formula,<br \/>\n\u03b8=2175\u03b8=2175 radian<br \/>\n=725=725radian<\/h3>\n<h2>Exercise 3.2<\/h2>\n<h2>1. Find the values of the other five trigonometric functions if cos x=-12cos x=-12 , xx lies in the third quadrant.<\/h2>\n<h3>Ans: Here given that, cos x=-12cos x=-12<br \/>\nTherefore we have,<br \/>\nsec x=1cos xsec x=1cos x<br \/>\n=1(-12)=1(-12)<br \/>\n=-2=-2<br \/>\nNow we know that,sin2x+cos2x=1sin2x+cos2x=1<br \/>\nTherefore we have, sin2x=1-cos2xsin2x=1-cos2x<br \/>\nSubstituting cos x=-12cos x=-12 in the formula, we obtain,<br \/>\nsin2x=1-(-12)2sin2x=1-(-12)2<br \/>\nsin2x=1-14sin2x=1-14<br \/>\n=34=34<br \/>\nsin x=\u00b13\u2013\u221a2sin x=\u00b132<br \/>\nSince xx lies in the 3rd3rdquadrant, the value of sinxsin\u2061x will be negative.<br \/>\nsin x=-3\u2013\u221a2sin x=-32<br \/>\nTherefore, cosec x=1sin xcosec x=1sin x<br \/>\n=1(-3\u2013\u221a2)=1(-32)<br \/>\n=-23\u2013\u221a=-23<br \/>\nHence ,<br \/>\ntan x=sin xcos xtan x=sin xcos x<br \/>\n=(-3\u2013\u221a2)(-12)=(-32)(-12)<br \/>\n=3\u2013\u221a=3<br \/>\nAnd<br \/>\ncot x=1tan xcot x=1tan x<br \/>\n=13\u2013\u221a=13<\/h3>\n<h2>2. Find the values of other five trigonometric functions if sin x=35sin x=35 , xx lies in second quadrant.<\/h2>\n<h3>Ans:<br \/>\nHere given that, sin x=35sin x=35<br \/>\nTherefore we have,<br \/>\ncosec x=1sin xcosec x=1sin x<br \/>\n=1(35)=1(35)<br \/>\n=53=53<br \/>\nNow we know that , sin2x+cos2x=1sin2x+cos2x=1<br \/>\nTherefore we have, cos2x=1-sin2xcos2x=1-sin2x<br \/>\nSubstituting sinx=35sin\u2061x=35 in the formula, we obtain,<br \/>\ncos2x=1-(35)2cos2x=1-(35)2<br \/>\ncos2x=1-925cos2x=1-925<br \/>\n=1625=1625<br \/>\ncos x=\u00b145cos x=\u00b145<br \/>\nSince xx lies in the 2nd2ndquadrant, the value of cosxcos\u2061x will be negative.<br \/>\ncos x=-45cos x=-45<br \/>\nTherefore, secx=1cosxsecx=1cos\u2061x<br \/>\n=1(-45)=1(-45)<br \/>\n=-54=-54<br \/>\nHence ,<br \/>\ntanx=sinxcosxtan\u2061x=sin\u2061xcos\u2061x<br \/>\n=(35)(-45)=(35)(-45)<br \/>\n=-34=-34<br \/>\nAnd<br \/>\ncotx=1tanxcot\u2061x=1tan\u2061x<br \/>\n=-43=-43<\/h3>\n<h2>3. Find the values of other five trigonometric functions if cot x=34cot x=34 , xx lies in third quadrant.<\/h2>\n<h3>Ans: Here given that, cotx=34cot\u2061x=34<br \/>\nTherefore we have,<br \/>\ntanx=1cotxtan\u2061x=1cot\u2061x<br \/>\n=1(34)=1(34)<br \/>\n=43=43<br \/>\nNow we know that ,<br \/>\nsec2x-tan2x=1sec2x-tan2x=1<br \/>\nTherefore we have, sec2x=1+tan2xsec2x=1+tan2x<br \/>\nSubstituting tan x=43tan x=43 in the formula, we obtain,<br \/>\nsec2x=1+(43)2sec2x=1+(43)2<br \/>\nsec2x=1+169sec2x=1+169<br \/>\n=259=259<br \/>\nsec x=\u00b153sec x=\u00b153<br \/>\nSince xx lies in the 3rd3rdquadrant, the value of secxsec\u2061x will be negative.<br \/>\nsec x=-53sec x=-53<br \/>\nTherefore, cosx=1secxcos\u2061x=1sec\u2061x<br \/>\n=1(-53)=1(-53)<br \/>\n=-35=-35<br \/>\nNow , tan x=sin xcos xtan x=sin xcos x<br \/>\nTherefore, sin x=tan xcos xsin x=tan xcos x<br \/>\nHence we have, sin x=43\u00d7(-35)sin x=43\u00d7(-35)<\/h3>\n<h3>=(-45)=(-45)<\/h3>\n<h3>And<br \/>\ncosec x=1sin xcosec x=1sin x<br \/>\n=-54=-54<\/h3>\n<h2>4. Find the values of other five trigonometric functions if sec x=135sec x=135 , xx lies in fourth quadrant.<\/h2>\n<h3>Ans: Here given that, secx=135sec\u2061x=135<br \/>\nTherefore we have,<br \/>\ncosx=1secxcos\u2061x=1sec\u2061x<br \/>\n=1(135)=1(135)<br \/>\n=513=513<br \/>\nNow we know that , sec2x-tan2x=1sec2x-tan2x=1<br \/>\nTherefore we have, tan2x=sec2x-1tan2x=sec2x-1<br \/>\nSubstituting<br \/>\nsec x=135sec x=135<br \/>\nin the formula, we obtain,<br \/>\ntan2x=(135)2-1tan2x=(135)2-1<br \/>\ntan2x=16925-1tan2x=16925-1<br \/>\n=14425=14425<\/h3>\n<h3>tanx=\u00b1125tanx=\u00b1125<br \/>\nSince xx lies in the 4th4th quadrant, the value of tanxtan\u2061x will be negative.<br \/>\ntan x=-125tan x=-125<br \/>\nTherefore,<br \/>\ncot x=1tan xcot x=1tan x<\/h3>\n<h3>=-512=-512<br \/>\nNow , tan x=sin xcos xtan x=sin xcos x<br \/>\nTherefore, sin x=tan xcos xsin x=tan xcos x<br \/>\nHence we have, sin x=513\u00d7(-125)sin x=513\u00d7(-125)<br \/>\n=(-1213)=(-1213)<br \/>\nAnd<br \/>\ncosec x=1sin xcosec x=1sin x<br \/>\n=-1312=-1312<\/h3>\n<h2>5. Find the values of other five trigonometric functions if tan x=-512tan x=-512 , xx lies in second quadrant.<\/h2>\n<h3>Ans: Here given that, tan x=-512tan x=-512<br \/>\nTherefore we have,<br \/>\ncot x=1tan xcot x=1tan x<br \/>\n=1(-512)=1(-512)<br \/>\n=-125=-125<br \/>\nNow we know that , sec2x-tan2x=1sec2x-tan2x=1<br \/>\nTherefore we have, sec2x=1+tan2xsec2x=1+tan2x<br \/>\nSubstituting tan x=-512tan x=-512 in the formula, we obtain,<br \/>\nsec2x=1+(-512)2sec2x=1+(-512)2<br \/>\nsec2x=1+25144sec2x=1+25144<br \/>\n=169144=169144<br \/>\nsec x=\u00b11312sec x=\u00b11312<br \/>\nSince xx lies in the 2nd2nd quadrant, the value of secxsec\u2061x will be negative.<br \/>\nsec x=-1312sec x=-1312<br \/>\nTherefore, cos x=1sec xcos x=1sec x<br \/>\n=-1213=-1213<br \/>\nNow , tan x=sin xcos xtan x=sin xcos x<br \/>\nTherefore, sin x=tan xcos xsin x=tan xcos x<br \/>\nHence we have, sin x=(-512)\u00d7(-1213)sin x=(-512)\u00d7(-1213)<br \/>\n=(513)=(513)<br \/>\nAnd<br \/>\ncosec x=1sin xcosec x=1sin x<br \/>\n=135=135<\/h3>\n<h2>6. Find the value of the trigonometric function sin765osin765o .<\/h2>\n<h3>Ans: We know that the values of sinxsin\u2061x repeat after an interval of 2\u03c02\u03c0 or 360\u2218360\u2218 .<br \/>\nTherefore we can write,<br \/>\nsin765o=sin(2\u00d7360o+45o)sin765o=sin(2\u00d7360o+45o)<br \/>\n=sin45o=sin45o<br \/>\n=12\u2013\u221a.=12.<\/h3>\n<h2>7. Find the value of the trigonometric function cosec(-1410o)cosec(-1410o)<\/h2>\n<h3>Ans: We know that the values of cosec xcosec x repeat after an interval of 2\u03c02\u03c0 or 360\u2218360\u2218 .<br \/>\nTherefore we can write,<br \/>\ncosec(-1410o)=cosec(-1410o+4\u00d7360o)cosec(-1410o)=cosec(-1410o+4\u00d7360o)<br \/>\n=cosec(-1410o+1440o)=cosec(-1410o+1440o)<br \/>\n=cosec30o=cosec30o<br \/>\n=2=2<\/h3>\n<h2>8. Find the value of the trigonometric function tan19\u03c03tan19\u03c03 .<\/h2>\n<h3>Ans: We know that the values of tan xtan x repeat after an interval of \u03c0\u03c0 or<br \/>\n180\u2218180\u2218<br \/>\n.<br \/>\nTherefore we can write,<br \/>\ntan19\u03c03=tan613\u03c0tan19\u03c03=tan613\u03c0<br \/>\n=tan(6\u03c0+\u03c03)=tan(6\u03c0+\u03c03)<\/h3>\n<h3>=tan\u03c03=tan\u03c03<br \/>\n=3\u2013\u221a=3<\/h3>\n<h2>9. Find the value of the trigonometric function sin(-11\u03c03)sin(-11\u03c03)<\/h2>\n<h3>Ans: We know that the values of sin xsin x repeat after an interval of 2\u03c02\u03c0 or 360\u2218360\u2218 .<br \/>\nTherefore we can write,<br \/>\nsin(-11\u03c03)=sin(-11\u03c03+2\u00d72\u03c0)sin(-11\u03c03)=sin(-11\u03c03+2\u00d72\u03c0)<br \/>\n=sin(\u03c03)=sin(\u03c03)<br \/>\n=3\u2013\u221a2=32<\/h3>\n<h2>10. Find the value of the trigonometric function cot(-15\u03c04)cot(-15\u03c04)<\/h2>\n<h3>Ans: We know that the values of cot xcot x repeat after an interval of \u03c0\u03c0 or<br \/>\n180\u2218180\u2218<br \/>\n.<br \/>\nTherefore we can write,<br \/>\ncot(-15\u03c04)=cot(-15\u03c04+4\u03c0)cot(-15\u03c04)=cot(-15\u03c04+4\u03c0)<br \/>\n=cot\u03c04=cot\u03c04<br \/>\n=1=1<\/h3>\n<h2>Exercise 3.3<\/h2>\n<h2>1. Prove that sin2\u03c06+cos2\u03c03-tan2\u03c04=-12sin2\u03c06+cos2\u03c03-tan2\u03c04=-12<\/h2>\n<h3>Ans: Substituting the values of<br \/>\nsin\u03c06,cos\u03c06,tan\u03c04sin\u03c06,cos\u03c06,tan\u03c04<br \/>\non left hand side,<br \/>\nsin2\u03c06+cos2\u03c03-tan2\u03c04=(12)2+(12)2-(1)2sin2\u03c06+cos2\u03c03-tan2\u03c04=(12)2+(12)2-(1)2<\/h3>\n<h3>=14+14-1=14+14-1<br \/>\n=\u221212=\u221212<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>2. Prove that 2sin2\u03c06+cosec27\u03c06cos2\u03c03=322sin2\u03c06+cosec27\u03c06cos2\u03c03=32<\/h2>\n<h3>Ans: Substituting the values of sin\u03c06,cosec7\u03c06,cos\u03c03sin\u03c06,cosec7\u03c06,cos\u03c03 on left hand side,<br \/>\nL.H.S.=2sin2\u03c06+cosec27\u03c06cos2\u03c03=2sin2\u03c06+cosec27\u03c06cos2\u03c03<br \/>\n=2(12)2+cosec2(\u03c0+\u03c06)(12)2=2(12)2+cosec2(\u03c0+\u03c06)(12)2<br \/>\n=2\u00d714+(-cosec\u03c06)2(14)=2\u00d714+(-cosec\u03c06)2(14)<br \/>\n=12+(-2)2(14)=12+(-2)2(14)<br \/>\nSince cosec xcosec x repeat its value after an interval of 2\u03c02\u03c0 ,<br \/>\nwe have, cosec7\u03c06=-cosec\u03c06cosec7\u03c06=-cosec\u03c06<br \/>\nL.H.S =12+44=12+44<br \/>\n=32=32<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>3. Prove that cot2\u03c06+cosec5\u03c06+3tan2\u03c06=6cot2\u03c06+cosec5\u03c06+3tan2\u03c06=6<\/h2>\n<h3>Ans: Substituting the values of cot\u03c06,cosec5\u03c06,tan\u03c06cot\u03c06,cosec5\u03c06,tan\u03c06 on left hand side,<br \/>\nL.H.S.=cot2\u03c06+cosec5\u03c06+3tan2\u03c06=cot2\u03c06+cosec5\u03c06+3tan2\u03c06<br \/>\n=(3\u2013\u221a)2+cosec(\u03c0-\u03c06)+3(13\u2013\u221a)2=(3)2+cosec(\u03c0-\u03c06)+3(13)2<br \/>\n=3+cosec\u03c06+3\u00d713=3+cosec\u03c06+3\u00d713<br \/>\nSince cosec xcosec x repeat its value after an interval of 2\u03c02\u03c0 ,<br \/>\nwe have, cosec5\u03c06=cosec\u03c06cosec5\u03c06=cosec\u03c06<br \/>\nL.H.S =3+2+1=3+2+1<br \/>\n=1=1<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>4. Prove that 2sin23\u03c04+2cos2\u03c04+2sec2\u03c03=102sin23\u03c04+2cos2\u03c04+2sec2\u03c03=10<\/h2>\n<h3>Ans:<br \/>\nSubstituting the values of sin3\u03c04,cos\u03c04,sec\u03c03sin3\u03c04,cos\u03c04,sec\u03c03 on left hand side,<br \/>\nL.H.S.=2sin23\u03c04+2cos2\u03c04+2sec2\u03c03=2sin23\u03c04+2cos2\u03c04+2sec2\u03c03<br \/>\n=2{sin(\u03c0-\u03c04)}2+2(12\u2013\u221a)2+2(2)2=2{sin(\u03c0-\u03c04)}2+2(12)2+2(2)2<br \/>\n=2{sin\u03c04}2+2\u00d712+8=2{sin\u03c04}2+2\u00d712+8<br \/>\nSince sin xsin x repeat its value after an interval of<br \/>\n2\u03c02\u03c0<br \/>\n,<br \/>\nwe have,<br \/>\nsin3\u03c04=sin\u03c04sin3\u03c04=sin\u03c04<\/h3>\n<h3>L.H.S =1+1+8=1+1+8<br \/>\n=10=10<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>5. Find the value of :<br \/>\n(i) sin75osin75o<\/h2>\n<h3>Ans: We have,<br \/>\nsin75o=sin(45o+30o)sin75o=sin(45o+30o)<br \/>\n=sin45ocos30o+cos45osin30o=sin45ocos30o+cos45osin30o<br \/>\nSince we know that, sin(x+y)=sin x cos y+cos x sin ysin(x+y)=sin x cos y+cos x sin y<br \/>\nTherefore we have,<br \/>\nSin75o=12\u2013\u221a\u00d73\u2013\u221a2+12\u2013\u221a\u00d712Sin75o=12\u00d732+12\u00d712<br \/>\nSin75o=3\u2013\u221a+122\u2013\u221aSin75o=3+122<\/h3>\n<h2>(ii) tan15otan15o<\/h2>\n<h3>Ans: We have,<br \/>\ntan15o=tan(45o-30o)tan15o=tan(45o-30o)<br \/>\n=tan45o-tan30o1+tan45otan30o=tan45o-tan30o1+tan45otan30o<br \/>\nSince we know, tan(x-y)=tan x-tan y1+tan x tan ytan(x-y)=tan x-tan y1+tan x tan y<br \/>\nTherefore we have,<br \/>\ntan15o=1-13\u2013\u221a1+1(13\u2013\u221a)tan15o=1-131+1(13)<br \/>\n=3\u2013\u221a-13\u2013\u221a3\u2013\u221a+13\u2013\u221a=3-133+13<br \/>\n=3\u2013\u221a-13\u2013\u221a+1=3-13+1<br \/>\n=(3\u2013\u221a-1)2(3\u2013\u221a+1)(3\u2013\u221a-1)=(3-1)2(3+1)(3-1)<br \/>\nFurther computing we have,<br \/>\ntan15o=3+1-23\u2013\u221a(3\u2013\u221a)2-(1)2tan15o=3+1-23(3)2-(1)2<\/h3>\n<h3>=4-23\u2013\u221a3-1=4-233-1<\/h3>\n<h3>=2-3\u2013\u221a=2-3<\/h3>\n<h2>6. Prove that cos(\u03c04-x)cos(\u03c04-y)-sin(\u03c04-x)sin(\u03c04-y)=sin(x+y)cos(\u03c04-x)cos(\u03c04-y)-sin(\u03c04-x)sin(\u03c04-y)=sin(x+y)<\/h2>\n<h3>Ans: We know that, cos(x+y)=cos xcos y-sin xsin ycos(x+y)=cos xcos y-sin xsin y<br \/>\ncos(\u03c04-x)cos(\u03c04-y)-sin(\u03c04-x)sin(\u03c04-y)=cos[\u03c04-x+\u03c04-y]cos(\u03c04-x)cos(\u03c04-y)-sin(\u03c04-x)sin(\u03c04-y)=cos[\u03c04-x+\u03c04-y]<br \/>\n=cos[\u03c02-(x+y)]=cos[\u03c02-(x+y)]<br \/>\n=sin(x+y)=sin(x+y)<br \/>\nL.H.S == R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>7. Prove that tan(\u03c04+x)tan(\u03c04-x)=(1+tanx1-tanx)2tan(\u03c04+x)tan(\u03c04-x)=(1+tanx1-tanx)2<\/h2>\n<h3>Ans: We know that ,tan(A+B)=tan A+tan B1-tan Atan Btan(A+B)=tan A+tan B1-tan Atan B<br \/>\nand tan(A-B)=tan A-tan B1+tan Atan Btan(A-B)=tan A-tan B1+tan Atan B<br \/>\nL.H.S.=tan(\u03c04+x)tan(\u03c04-x)=tan(\u03c04+x)tan(\u03c04-x)<br \/>\nUsing the above formula,<br \/>\nL.H.S=\u239b\u239d\u239ctan\u03c04+tanx1-tan\u03c04tanx\u239e\u23a0\u239ftan\u03c04-tanx1+tan\u03c04tanxL.H.S=(tan\u03c04+tanx1-tan\u03c04tanx)tan\u03c04-tanx1+tan\u03c04tanx<br \/>\n=(1+tan x1-tan x)(1-tan x1+tan x)=(1+tan x1-tan x)(1-tan x1+tan x) substituting$tan\u03c04=1$substituting$tan\u03c04=1$<br \/>\n=(1+tan x1-tan x)2=(1+tan x1-tan x)2<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>8. Prove that cos(\u03c0+x)cos(-x)sin(\u03c0-x)cos(\u03c02+x)=cot2xcos(\u03c0+x)cos(-x)sin(\u03c0-x)cos(\u03c02+x)=cot2x<\/h2>\n<h3>Ans: Observe that cos xcos x repeats same value after an interval 2\u03c02\u03c0<br \/>\nand sin xsin x repeats same value after an interval 2\u03c02\u03c0.<br \/>\nL.H.S.=cos(\u03c0+x)cos(-x)sin(\u03c0-x)cos(\u03c02+x)=cos(\u03c0+x)cos(-x)sin(\u03c0-x)cos(\u03c02+x)<br \/>\n=[-cos x][cos x](sin x)(-sin x)=[-cos x][cos x](sin x)(-sin x)<br \/>\n=-cos2x-sin2x=-cos2x-sin2x<br \/>\n=cot2x=cot2x<br \/>\nHence proved.<\/h3>\n<h2>9. Prove that,<br \/>\nCos(3\u03c02+x)Cos(2\u03c0+x)[cot(3\u03c02-x)+cot(2\u03c0+x)]=1Cos(3\u03c02+x)Cos(2\u03c0+x)[cot(3\u03c02-x)+cot(2\u03c0+x)]=1<\/h2>\n<h3>Ans: We know that cot xcot x repeats same value after an interval 2\u03c02\u03c0 .<br \/>\nL.H.S.=Cos(3\u03c02+x)Cos(2\u03c0+x)[cot(3\u03c02\u2212x)+cot(2\u03c0+x)]=Cos(3\u03c02+x)Cos(2\u03c0+x)[cot(3\u03c02\u2212x)+cot(2\u03c0+x)]<br \/>\n=sin x cos x[tan x+cot x]=sin x cos x[tan x+cot x]<br \/>\nSubstituting tan x=sin xcos xtan x=sin xcos x and<br \/>\ncot x=cos xsin xcot x=cos xsin x ,<br \/>\nL.H.S=sin xcos x(sin xcos x+cos xsin x)L.H.S=sin xcos x(sin xcos x+cos xsin x)<br \/>\n=(sin x cos x)[sin2x+cos2xsin x cos x]=(sin x cos x)[sin2x+cos2xsin x cos x]<br \/>\n=1=1<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>10. Prove that sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=cos xsin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=cos x<\/h2>\n<h3>Ans: We know that , cos(x-y)=cosxcosy+sinxsinycos(x-y)=cosxcosy+sinxsiny<br \/>\nL.H.S.=sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x<br \/>\n=cos[(n+1)x-(n+2)x]=cos[(n+1)x-(n+2)x]<br \/>\n=cos(-x)=cos(-x)<br \/>\n=cosx=cosx<br \/>\n== R.H.S.<\/h3>\n<h2>11. Prove that cos(3\u03c04+x)-cos(3\u03c04-x)=-2\u2013\u221asinxcos(3\u03c04+x)-cos(3\u03c04-x)=-2sinx<\/h2>\n<h3>Ans: We know that , cos A-cos B=-2sin(A+B2).sin(A-B2)cos A-cos B=-2sin(A+B2).sin(A-B2)<br \/>\n\u2234\u2234 L.H.S.=cos(3\u03c04+x)-cos(3\u03c04-x)=cos(3\u03c04+x)-cos(3\u03c04-x)<br \/>\n=-2sin\u23a7\u23a9\u23a8\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa(3\u03c04+x)+(3\u03c04-x)2\u23ab\u23ad\u23ac\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa.sin\u23a7\u23a9\u23a8\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa(3\u03c04+x)-(3\u03c04-x)2\u23ab\u23ad\u23ac\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa=-2sin{(3\u03c04+x)+(3\u03c04-x)2}.sin{(3\u03c04+x)-(3\u03c04-x)2}<br \/>\n=-2sin(3\u03c04)sin x=-2sin(3\u03c04)sin x<br \/>\nSince sin xsin x repeats the same value after an interval 2\u03c02\u03c0 ,<br \/>\nwe have, sin(3\u03c04)=sin(\u03c0-\u03c04)sin(3\u03c04)=sin(\u03c0-\u03c04)<br \/>\ntherefore,<br \/>\nL.H.S=-2sin\u03c04sin xL.H.S=-2sin\u03c04sin x<br \/>\n=-2\u00d712\u2013\u221a\u00d7sinx=-2\u00d712\u00d7sinx<br \/>\n=-2\u2013\u221asin x=-2sin x<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>12. Prove that sin26x-sin24x=sin 2x sin 10xsin26x-sin24x=sin 2x sin 10x<\/h2>\n<h3>Ans: We know that,sinA+sinB=2sin(A+B2)cos(A-B2)sinA+sinB=2sin(A+B2)cos(A-B2)<br \/>\nAnd sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)<br \/>\n\u2234\u2234 L.H.S.=sin26x-sin24xa=sin26x-sin24xa<br \/>\n=(sin 6x+sin 4x)(sin 6x-sin 4x)=(sin 6x+sin 4x)(sin 6x-sin 4x)<br \/>\n=[2sin(6x+4&#215;2)cos(6x-4&#215;2)][2cos(6x+4&#215;2).sin(6x-4&#215;2)]=[2sin(6x+4&#215;2)cos(6x-4&#215;2)][2cos(6x+4&#215;2).sin(6x-4&#215;2)]<br \/>\n=(2sin 5x cos x)(2cos 5x sin x)=(2sin 5x cos x)(2cos 5x sin x)<br \/>\nNow we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x ,<br \/>\nTherefore we have,<br \/>\nL.H.S=(2sin 5x cos 5x)(2sin x cos x)L.H.S=(2sin 5x cos 5x)(2sin x cos x)<br \/>\n=sin 10x sin 2x=sin 10x sin 2x<br \/>\n== R.H.S.<\/h3>\n<h2>13. Prove that cos22x-cos26x=sin 4x sin 8xcos22x-cos26x=sin 4x sin 8x<\/h2>\n<h3>Ans: We know that,<br \/>\ncos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nAnd cos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)<br \/>\nL.H.S.=cos22x-cos26x=cos22x-cos26x<br \/>\n=(cos 2x+cos 6x)(cos 2x-6x)=(cos 2x+cos 6x)(cos 2x-6x)<br \/>\n=[2cos(2x+6&#215;2)cos(2x-6&#215;2)][-2sin(2x+6&#215;2)sin(2x-6&#215;2)]=[2cos(2x+6&#215;2)cos(2x-6&#215;2)][-2sin(2x+6&#215;2)sin(2x-6&#215;2)]<br \/>\nFurther computing, we have,<br \/>\nL.H.S=[2cos 4x cos(-2x)][-2sin 4xsin(-2x)]L.H.S=[2cos 4x cos(-2x)][-2sin 4xsin(-2x)]<br \/>\n=[2cos 4x cos 2x][-2sin 4x(-sin 2x)]=[2cos 4x cos 2x][-2sin 4x(-sin 2x)]<br \/>\n=(2sin 4x cos 4x)(2sin 2xcos 2x)=(2sin 4x cos 4x)(2sin 2xcos 2x)<br \/>\nNow we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x<br \/>\nTherefore we have,<br \/>\nL.H.S=sin 8x sin 4xL.H.S=sin 8x sin 4x<br \/>\n== R.H.S.<br \/>\n.Hence proved.<\/h3>\n<h2>14. Prove that sin 2x+2sin 4x+sin6=4cos2xsin 4xsin 2x+2sin 4x+sin6=4cos2xsin 4x<\/h2>\n<h3>Ans: We know that,<br \/>\nsin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)<br \/>\nL.H.S.=sin 2x+2sin 4x+sin 6x=sin 2x+2sin 4x+sin 6x<br \/>\n. =[sin 2x+sin 6x]+2sin 4x=[sin 2x+sin 6x]+2sin 4x<br \/>\n=[2sin(2x+6&#215;2)cos(2x-6&#215;2)]+2sin4x=[2sin(2x+6&#215;2)cos(2x-6&#215;2)]+2sin4x<br \/>\n=2sin 4xcos(-2x)+2sin 4x=2sin 4xcos(-2x)+2sin 4x<br \/>\nFurther computing,<br \/>\nWe have, L.H.S=2sin 4x cos 2x+2sin 4xL.H.S=2sin 4x cos 2x+2sin 4x<br \/>\n=2sin 4x(cos 2x+1)=2sin 4x(cos 2x+1)<br \/>\nNow we know that, cos 2x+1=2cos2xcos 2x+1=2cos2x<br \/>\nTherefore we have,<br \/>\nL.H.S=2sin 4x(2cos2x)L.H.S=2sin 4x(2cos2x)<br \/>\n=4cos2xsin 4x=4cos2xsin 4x<br \/>\n= R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>15. Prove that cot 4x(sin 5x+sin 3x)=cot x(sin 5x-sin 3x)cot 4x(sin 5x+sin 3x)=cot x(sin 5x-sin 3x)<\/h2>\n<h3>Ans: We know that, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)<br \/>\nL.H.S.=cot 4x(sin 5x+sin 3x)=cot 4x(sin 5x+sin 3x)<br \/>\n=cot 4xsin 4x[2sin(5x+3&#215;2)cos(5x+3&#215;2)]=cot 4xsin 4x[2sin(5x+3&#215;2)cos(5x+3&#215;2)]<br \/>\n=(cos 4xsin 4x)[2sin 4x cos x]=(cos 4xsin 4x)[2sin 4x cos x]<br \/>\n=2cos 4x cos x=2cos 4x cos x<br \/>\nNow also ,we know that, sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)<br \/>\nR.H.S.=cot x(sin 5x-sin 3x)=cot x(sin 5x-sin 3x)<br \/>\n=cos xsin x[2cos(5x+3&#215;2)sin(5x-3&#215;2)]=cos xsin x[2cos(5x+3&#215;2)sin(5x-3&#215;2)]<\/h3>\n<h3>=cos xsin x[2cos 4x sin x]=cos xsin x[2cos 4x sin x]<br \/>\n=2cos 4x cos x=2cos 4x cos x<br \/>\nTherefore , we can conclude that,<br \/>\nL.H.S.=R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>16. Prove that cos 9x-cos 5xsin 17x-sin 3x=-sin 2xcos 10xcos 9x-cos 5xsin 17x-sin 3x=-sin 2xcos 10x<\/h2>\n<h3>Ans: We know that,<br \/>\ncos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)<br \/>\nAnd sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)<br \/>\nL.H.S.=cos 9x-cos 5xsin 17x-sin 3x=cos 9x-cos 5xsin 17x-sin 3x<\/h3>\n<h3>=-2sin(9x+5&#215;2).sin(9x-5&#215;2)2cos(17x+3&#215;2).sin(17x-3&#215;2)=-2sin(9x+5&#215;2).sin(9x-5&#215;2)2cos(17x+3&#215;2).sin(17x-3&#215;2)<br \/>\n(following the formula)<\/h3>\n<h3>=-2sin 7x.sin 2x2cos 10x.sin 7x=-2sin 7x.sin 2x2cos 10x.sin 7x<br \/>\n=-sin 2xcos 10x=-sin 2xcos 10x<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>17. Prove that:sin 5x+sin 3xcos 5x+cos 3x=tan 4xsin 5x+sin 3xcos 5x+cos 3x=tan 4x<\/h2>\n<h3>Ans:<br \/>\nWe know that<br \/>\nsin A+sin B=2sin(A+B2)cos(A-B2),sin A+sin B=2sin(A+B2)cos(A-B2),<br \/>\ncos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nNow , L.H.S.=sin 5x+sin 3xcos 5x+cos 3x=sin 5x+sin 3xcos 5x+cos 3x<br \/>\n=2sin(5x+3&#215;2)cos(5x-3&#215;2)2cos(5x+3&#215;2)cos(5x-3&#215;2)=2sin(5x+3&#215;2)cos(5x-3&#215;2)2cos(5x+3&#215;2)cos(5x-3&#215;2) (using the formula)<br \/>\n=2sin(5x+3&#215;2)cos(5x-3&#215;2)2cos(5x+3&#215;2)cos(5x-3&#215;2)=2sin(5x+3&#215;2)cos(5x-3&#215;2)2cos(5x+3&#215;2)cos(5x-3&#215;2)<br \/>\n=2sin 4x cos x2cos 4x cos x=2sin 4x cos x2cos 4x cos x<br \/>\nFurther computing we have,<br \/>\nL.H.S=tan 4xL.H.S=tan 4x<br \/>\n== R.H.S.<\/h3>\n<h2>18. Prove that<br \/>\nsin x-sin ycos x+cos y=tanx-y2sin x-sin ycos x+cos y=tanx-y2<\/h2>\n<h3>Ans: We know that,<br \/>\nsin A-sin B=2cos(A+B2)sin(A-B2),sin A-sin B=2cos(A+B2)sin(A-B2),<br \/>\n.cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nL.H.S.<br \/>\n=sin x-sin ycosx+cosy=sin x-sin ycosx+cosy<br \/>\n=2cos(x+y2).sin(x-y2)2cos(x+y2).cos(x-y2)=2cos(x+y2).sin(x-y2)2cos(x+y2).cos(x-y2)<br \/>\n=sin(x-y2)cos(x-y2)=sin(x-y2)cos(x-y2)<br \/>\n=tan(x-y2)=tan(x-y2)<br \/>\nTherefore L.H.S=R.H.SL.H.S=R.H.S<br \/>\nHence proved.<\/h3>\n<h2>19. Prove that sin x+sin 3xcos x+cos 3x=tan 2xsin x+sin 3xcos x+cos 3x=tan 2x<\/h2>\n<h3>Ans: We know that<br \/>\nsin A+sin B=2sin(A+B2)cos(A+B2),sin A+sin B=2sin(A+B2)cos(A+B2),<br \/>\n.<br \/>\ncos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nNow , L.H.S.=sinx+sin3xcos x+cos 3x=sinx+sin3xcos x+cos 3x<br \/>\n=2sin(x+3&#215;2)cos(x-3&#215;2)2cos(x+3&#215;2)cos(x-3&#215;2)=2sin(x+3&#215;2)cos(x-3&#215;2)2cos(x+3&#215;2)cos(x-3&#215;2) (using the formula)<br \/>\n=sin 2xcos 2x=sin 2xcos 2x<br \/>\n=tan 2x=tan 2x<br \/>\nTherefore L.H.S== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>20. Prove that sin x-sin 3xsin2x-cos2x=2sin xsin x-sin 3xsin2x-cos2x=2sin x<\/h2>\n<h3>Ans: We know that,<br \/>\nsin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)<br \/>\nAnd<br \/>\ncos2A-sin2A=cos 2Acos2A-sin2A=cos 2A<br \/>\nL.H.S.=sin x-sin 3xsin2x-cos2x=sin x-sin 3xsin2x-cos2x<\/h3>\n<h3>=2cos(x+3&#215;2)sin(x-3&#215;2)-cos2x=2cos(x+3&#215;2)sin(x-3&#215;2)-cos2x<br \/>\n=2cos2xsin(-x)-cos 2x=2cos2xsin(-x)-cos 2x<br \/>\n=-2\u00d7(-sinx)=-2\u00d7(-sinx)<br \/>\nTherefore , we have,<br \/>\nL.H.S=2sin xL.H.S=2sin x<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>21. Prove that cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cot 3xcos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cot 3x<\/h2>\n<h3>Ans: We know that,<br \/>\ncos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nAnd, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)<br \/>\nNow, L.H.S.=cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x<\/h3>\n<h3>=(cos 4x+cos 2x)+cos 3x(sin4x+sin2x)+sin 3x=(cos 4x+cos 2x)+cos 3x(sin4x+sin2x)+sin 3x<\/h3>\n<h3>=2cos(4x+2&#215;2)cos(4x-2&#215;2)+cos3x2sin(4x+2&#215;2)cos(4x-2&#215;2)+sin 3x=2cos(4x+2&#215;2)cos(4x-2&#215;2)+cos3x2sin(4x+2&#215;2)cos(4x-2&#215;2)+sin 3x<br \/>\n(using the formulas)<\/h3>\n<h3>=2cos 3x cos x+cos 3x2sin 3x cos x+sin 3x=2cos 3x cos x+cos 3x2sin 3x cos x+sin 3x<br \/>\nFurther computing, we obtain,<br \/>\nL.H.S=cos 3x(2cos x+1)sin 3x(2cos x+1)=cos 3x(2cos x+1)sin 3x(2cos x+1)<\/h3>\n<h3>=cot 3x=cot 3x<\/h3>\n<h3>== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>22. Prove that<br \/>\ncot x cot 2x-cot 2x cot 3x-cot 3x cot x=1cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1<\/h2>\n<h3>Ans:<br \/>\nWe know that,<br \/>\ncot(A+B)=cotAcotB-1cot A+cot Bcot(A+B)=cotAcotB-1cot A+cot B<br \/>\nNow , L.H.S.=cot xcot 2x-cot 2x cot 3x-cot 3x cot x=cot xcot 2x-cot 2x cot 3x-cot 3x cot x<\/h3>\n<h3>=cot x cot 2x-cot 3x(cot 2x+cot x)=cot x cot 2x-cot 3x(cot 2x+cot x)<\/h3>\n<h3>=cot x cot 2x-cot(2x+x)(cot 2x+cot x)=cot x cot 2x-cot(2x+x)(cot 2x+cot x)<\/h3>\n<h3>=cot x cot 2x-[cot 2x cot x-1cot x+cot 2x](cot 2x+cot x)=cot x cot 2x-[cot 2x cot x-1cot x+cot 2x](cot 2x+cot x)<br \/>\nFurther computing we obtain,<br \/>\nL.H.S=cot x cot 2x-(cot 2x cot x-1)L.H.S=cot x cot 2x-(cot 2x cot x-1)<\/h3>\n<h3>=1=1<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>23. Prove that tan 4x=4tan x(1-tan2x)1-6tan2x+tan4xtan 4x=4tan x(1-tan2x)1-6tan2x+tan4x<\/h2>\n<h3>Ans: We know that tan 2A=2tan A1-tan2Atan 2A=2tan A1-tan2A<br \/>\nL.H.S.=tan 4x=tan 4x<br \/>\n=tan2(2x)=tan2(2x)<\/h3>\n<h3>=2tan 2&#215;1-tan2(2x)=2tan 2&#215;1-tan2(2x)<br \/>\nusingtheformulausingtheformula<br \/>\n=(4tan x1-tan2x)[1-4tan2x(1-tan2x)2]=(4tan x1-tan2x)[1-4tan2x(1-tan2x)2]<br \/>\nFurther computing, we obtain,<br \/>\nL.H.S =(4tan x1-tan2x)[(1-tan2x)24tan2x(1-tan2x)2]=(4tan x1-tan2x)[(1-tan2x)24tan2x(1-tan2x)2]<br \/>\n=4tan x(1-tan2x)1+tan4x-2tan2x-4tan2x=4tan x(1-tan2x)1+tan4x-2tan2x-4tan2x<br \/>\n=4tan x(1-tan2x)1-6tan2x+tan4x=4tan x(1-tan2x)1-6tan2x+tan4x<br \/>\n== R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>24. Prove that cos 4x=1-8sin2xcos2xcos 4x=1-8sin2xcos2x<\/h2>\n<h3>Ans: We know that, cos 2x=1-2sin2xcos 2x=1-2sin2x<br \/>\nAnd sin 2x=2sin x cos xsin 2x=2sin x cos x<br \/>\nL.H.S.<br \/>\n=cos 4x=cos 4x<br \/>\n=cos 2(2x)=cos 2(2x)<br \/>\n=1-2sin22x=1-2sin22x<br \/>\n=1-2(2sin x cos x)2=1-2(2sin x cos x)2<br \/>\nFurther computing we get,<br \/>\nL.H.S=1-8sin2xcos2x=1-8sin2xcos2x<br \/>\n==R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>25. Prove that cos 6x=32xcos6x-48cos4x+18cos2x-1cos 6x=32xcos6x-48cos4x+18cos2x-1<\/h2>\n<h3>Ans: We know that, cos 3A=4cos3A-3cosAcos 3A=4cos3A-3cosA<br \/>\nand cos 2x=1-2sin2xcos 2x=1-2sin2x<br \/>\nL.H.S.=cos 6x=cos 6x<br \/>\n=cos 3(2x)=cos 3(2x)<\/h3>\n<h2>=4cos32x-3cos 2x=4cos32x-3cos 2x<\/h2>\n<h3>=4[(2cos2x-1)3-3(2cos2x-1)]=4[(2cos2x-1)3-3(2cos2x-1)]<br \/>\nFurther computing,<br \/>\nL.H.S=4[(2cos2x)3-(1)3-3(2cos2x)]-6cos2x+3=4[(2cos2x)3-(1)3-3(2cos2x)]-6cos2x+3<br \/>\n=4[(2cos2x)3-(1)3-3(2cos2x)2+3(2cos2x)]-6cos2x+3=4[(2cos2x)3-(1)3-3(2cos2x)2+3(2cos2x)]-6cos2x+3<br \/>\n=4[8cos6x-1-12cos4x+6cos2x]-6cos2x+3=4[8cos6x-1-12cos4x+6cos2x]-6cos2x+3<br \/>\n=32cos6x-48cos4x+18cos2x-1=32cos6x-48cos4x+18cos2x-1<br \/>\nTherefore we have,<br \/>\nL.H.S == R.H.S.<br \/>\nHence proved.<\/h3>\n<h2>Exercise 3.4<\/h2>\n<h2>1. Find the principal and general solutions of the<br \/>\ntan x=3\u2013\u221atan x=3 .<\/h2>\n<h3>Ans: Here given that,<br \/>\ntan x=3\u2013\u221atan x=3<br \/>\nWe know that tan\u03c03=3\u2013\u221atan\u03c03=3<br \/>\nand tan(4\u03c03)=tan(\u03c0+\u03c03)=tan\u03c03=3\u2013\u221atan(4\u03c03)=tan(\u03c0+\u03c03)=tan\u03c03=3<br \/>\nTherefore, the principal solutions are<br \/>\nx=\u03c03x=\u03c03<br \/>\nand<br \/>\n4\u03c034\u03c03 .<br \/>\nNow,<br \/>\ntan x=tan\u03c03tan x=tan\u03c03<br \/>\nWhich implies,<br \/>\nx=n\u03c0+\u03c03x=n\u03c0+\u03c03 , where<br \/>\nn\u2208Zn\u2208Z<br \/>\nTherefore, the general solution is<br \/>\nx=n\u03c0+\u03c03x=n\u03c0+\u03c03<br \/>\n, where n\u2208Zn\u2208Z .<\/h3>\n<h2>2. Find the principal and general solutions of the equation secx=2secx=2<\/h2>\n<h3>Ans: Here it is given that,<br \/>\nsec x=2sec x=2<br \/>\nNow we know that<br \/>\nsec\u03c03=2sec\u03c03=2 and<br \/>\nsec5\u03c03=sec(2\u03c0-\u03c03)sec5\u03c03=sec(2\u03c0-\u03c03)<br \/>\n=sec\u03c03=sec\u03c03<br \/>\n=2=2<br \/>\nTherefore, the principal solutions arex=\u03c03x=\u03c03 and<br \/>\n5\u03c035\u03c03<br \/>\n.<br \/>\nNow, sec x=sec\u03c03sec x=sec\u03c03<br \/>\nand we know ,<br \/>\nsecx=1cosxsec\u2061x=1cos\u2061x<br \/>\nTherefore , we have,<br \/>\ncos x=cos\u03c03cos x=cos\u03c03<br \/>\nWhich implies,<br \/>\nx=2n\u03c0\u00b1\u03c03x=2n\u03c0\u00b1\u03c03 , where n\u2208Zn\u2208Z .<br \/>\nTherefore, the general solution is x=2n\u03c0\u00b1\u03c03x=2n\u03c0\u00b1\u03c03 , where n\u2208Zn\u2208Z .<\/h3>\n<h2>3. Find the principal and general solutions of the equation cot x=-3\u2013\u221acot x=-3<\/h2>\n<h3>Ans: Here it is given that,<br \/>\ncot x=-3\u2013\u221acot x=-3<br \/>\nNow we know that cot\u03c06=3\u2013\u221acot\u03c06=3<br \/>\nAnd<br \/>\ncot(\u03c0-\u03c06)=-cot\u03c06cot(\u03c0-\u03c06)=-cot\u03c06<\/h3>\n<h3>=-3\u2013\u221a=-3<br \/>\nand cot(2\u03c0-\u03c06)=-cot\u03c06cot(2\u03c0-\u03c06)=-cot\u03c06<\/h3>\n<h3>=-3\u2013\u221a=-3<br \/>\nTherefore we have,<br \/>\ncot5\u03c06=-3\u2013\u221acot5\u03c06=-3<br \/>\nand cot11\u03c06=-3\u2013\u221acot11\u03c06=-3<br \/>\nTherefore, the principal solutions are x=5\u03c06x=5\u03c06 and 11\u03c0611\u03c06.<br \/>\n.Now, cot x=cot5\u03c06cot x=cot5\u03c06<br \/>\nAnd we know<br \/>\ncot x=1tan xcot x=1tan x<br \/>\nTherefore we have,<br \/>\ntan x=tan5\u03c06tan x=tan5\u03c06<br \/>\nWhich implies,<br \/>\nx=n\u03c0+5\u03c06x=n\u03c0+5\u03c06 , where n\u2208Zn\u2208Z<br \/>\nTherefore, the general solution is x=n\u03c0+5\u03c06x=n\u03c0+5\u03c06 , where n\u2208Zn\u2208Z .<\/h3>\n<h2>4. Find the general solution of cosec x=-2cosec x=-2<\/h2>\n<h3>Ans: Here it is given that,<br \/>\ncosec x=-2cosec x=-2<br \/>\nNow we know that<br \/>\ncosec\u03c06=2cosec\u03c06=2<br \/>\nand<br \/>\ncosec(\u03c0+\u03c06)=-cosec\u03c06acosec(\u03c0+\u03c06)=-cosec\u03c06a<br \/>\n=-2=-2<br \/>\nand<br \/>\ncosec(2\u03c0-\u03c06)=-cosec\u03c06cosec(2\u03c0-\u03c06)=-cosec\u03c06<\/h3>\n<h3>=-2=-2<br \/>\ntherefore we have,<br \/>\ncosec7\u03c06=-2cosec7\u03c06=-2<br \/>\nand cosec11\u03c06=-2cosec11\u03c06=-2<br \/>\nHence , the principal solutions arex=7\u03c06x=7\u03c06 and 11\u03c0611\u03c06.<br \/>\nNow, cosec x=cosec7\u03c06cosec x=cosec7\u03c06<br \/>\nAnd we know,<br \/>\ncosec x=1sin xcosec x=1sin x<br \/>\nTherefore , we have,<br \/>\nsin x=sin7\u03c06sin x=sin7\u03c06<br \/>\nWhich implies,<br \/>\nx=n\u03c0+(-1)n7\u03c06x=n\u03c0+(-1)n7\u03c06<br \/>\n,where n\u2208Zn\u2208Z.<br \/>\nTherefore, the general solution is x=n\u03c0+(-1)n7\u03c06x=n\u03c0+(-1)n7\u03c06 ,where n\u2208Zn\u2208Z.<\/h3>\n<h2>5. Find the general solution of the equation cos 4x=cos 2xcos 4x=cos 2x<\/h2>\n<h3>Ans: Here it is given that, cos 4x=cos 2xcos 4x=cos 2x<br \/>\nWhich implies,<br \/>\ncos 4x-cos 2x=0cos 4x-cos 2x=0<br \/>\nNow we know that, cos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)<br \/>\nTherefore we have,<br \/>\n-2sin(4x+2&#215;2)sin(4x-2&#215;2)=0-2sin(4x+2&#215;2)sin(4x-2&#215;2)=0<br \/>\nsin 3x sin x=0sin 3x sin x=0<br \/>\nHence we have, sin 3x=0sin 3x=0<br \/>\nOr, sin x=0sin x=0<br \/>\nTherefore, 3x=n\u03c03x=n\u03c0<br \/>\nor x=n\u03c0x=n\u03c0 ,where n\u2208Zn\u2208Z<br \/>\ntherefore,<br \/>\nx=n\u03c03x=n\u03c03<\/h3>\n<h3>or x=n\u03c0x=n\u03c0 ,where n\u2208Zn\u2208Z.<\/h3>\n<h2>6. Find the general solution of the equation cos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0.<\/h2>\n<h3>Ans: Here given that,<br \/>\ncos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0<br \/>\nNow we know that, cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)<br \/>\nTherefore cos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0 implies<br \/>\n2cos(3x+x2)cos(3x-x2)-cos 2x=02cos(3x+x2)cos(3x-x2)-cos 2x=0<br \/>\n2cos 2x cos x-cos 2x=02cos 2x cos x-cos 2x=0<br \/>\ncos 2x(2cos x-1)=0cos 2x(2cos x-1)=0<br \/>\nHence we have,<br \/>\nEither cos 2x=0cos 2x=0<br \/>\nOr cos x=12cos x=12<br \/>\nWhich in turn implies that,<br \/>\nEither 2x=(2n+1)\u03c022x=(2n+1)\u03c02<br \/>\nOr, cos x=cos\u03c03cos x=cos\u03c03 , where n\u2208Zn\u2208Z<br \/>\nTherefore,<br \/>\nEither x=(2n+1)\u03c04x=(2n+1)\u03c04<br \/>\nOr, x=2n\u03c0\u00b1\u03c03x=2n\u03c0\u00b1\u03c03 ,where<br \/>\nn\u2208Zn\u2208Z<br \/>\n.<\/h3>\n<h2>7. Find the general solution of the equation<br \/>\nsin 2x+cos x=0sin 2x+cos x=0<br \/>\n.<\/h2>\n<h3>Ans: Here it is given that,<br \/>\nsin 2x cos x=0sin 2x cos x=0<br \/>\nNow we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x<br \/>\nTherefore we have,<br \/>\n2sin x cos x+cos x=02sin x cos x+cos x=0<br \/>\nWhich implies,<\/h3>\n<h3>cos x(2sin x+1)=0cos x(2sin x+1)=0<br \/>\nTherefore we have,<br \/>\nEither cos x=0cos x=0<br \/>\nOr, sin x=-12sin x=-12<br \/>\nHence we have,<\/h3>\n<h3>x=(2n+1)\u03c02x=(2n+1)\u03c02<br \/>\n, where n\u2208Zn\u2208Z .<br \/>\nEither<br \/>\nOr,<br \/>\nsin x=-12sin x=-12<br \/>\n=-sin\u03c06=-sin\u03c06<br \/>\n=sin(\u03c0-7\u03c06)=sin(\u03c0-7\u03c06)<br \/>\n=sin7\u03c06=sin7\u03c06<br \/>\nWhich implies<br \/>\nx=n\u03c0+(-1)n7\u03c06x=n\u03c0+(-1)n7\u03c06 , where n\u2208Zn\u2208Z<br \/>\nTherefore, the general solution is<br \/>\n(2n+1)\u03c02(2n+1)\u03c02<br \/>\nor<br \/>\nn\u03c0+(-1)n7\u03c06,n\u2208Zn\u03c0+(-1)n7\u03c06,n\u2208Z<br \/>\n.<\/h3>\n<h2>8. Find the general solution of the equation<br \/>\nsec22x=1-tan 2xsec22x=1-tan 2x<\/h2>\n<h3>Ans: Here given that ,<br \/>\nsec22x=1-tan 2xsec22x=1-tan 2x<br \/>\nNow we know that, sec2x-tan2x=1sec2x-tan2x=1<br \/>\nTherefore we have,<\/h3>\n<h3>sec22x=1-tan 2xsec22x=1-tan 2x<br \/>\nimplies<\/h3>\n<h3>1+tan22x=1-tan 2&#215;1+tan22x=1-tan 2x<\/h3>\n<h3>tan22x+tan 2x=0tan22x+tan 2x=0<\/h3>\n<h3>tan 2x(tan 2x+1)=0tan 2x(tan 2x+1)=0<br \/>\nHence either tan 2x=0tan 2x=0<br \/>\nOr, tan 2x=-1tan 2x=-1<br \/>\nWhich implies either<br \/>\nx=n\u03c02x=n\u03c02<br \/>\n, where n\u2208Zn\u2208Z ,<br \/>\nOr, tan 2x=-1tan 2x=-1<br \/>\n=-tan\u03c04=-tan\u03c04<br \/>\n=tan(\u03c0-\u03c04)=tan(\u03c0-\u03c04)<br \/>\n=tan3\u03c04=tan3\u03c04<br \/>\nWhich in turn implies that,<br \/>\n2x=n\u03c0+3\u03c04,2x=n\u03c0+3\u03c04, where n\u2208Zn\u2208Z<br \/>\ni.e, x=n\u03c02+3\u03c08,x=n\u03c02+3\u03c08, where n\u2208Zn\u2208Z.<br \/>\nTherefore, the general solution is n\u03c02n\u03c02 or n\u03c02+3\u03c08,n\u2208Zn\u03c02+3\u03c08,n\u2208Z.<\/h3>\n<h2>9. Find the general solution of the equation<br \/>\nsin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0<\/h2>\n<h3>Ans:<br \/>\nHere given that ,<br \/>\nsin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0<br \/>\nNow we know that, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)<br \/>\nTherefore ,<\/h3>\n<h3>sin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0<\/h3>\n<h3>(sin x+sin 3x)+sin 5x=0(sin x+sin 3x)+sin 5x=0<\/h3>\n<h3>[2sin(x+5&#215;2)cos(x-5&#215;2)]+sin 3x=0[2sin(x+5&#215;2)cos(x-5&#215;2)]+sin 3x=0<\/h3>\n<h3>2sin 3x cos (-2x)+sin 3x=02sin 3x cos (-2x)+sin 3x=0<br \/>\nSimplifying we get,<br \/>\n2sin 3xcos 2x+sin 3x=02sin 3xcos 2x+sin 3x=0<\/h3>\n<h3>sin 3x(2cos 2x+1)=0sin 3x(2cos 2x+1)=0<br \/>\nHence either sin 3x=0sin 3x=0<br \/>\nOr, cos 2x=-12cos 2x=-12<br \/>\nWhich implies 3x=n\u03c03x=n\u03c0 , where n\u2208Zn\u2208Z<br \/>\nOr, cos 2x=-12cos 2x=-12<br \/>\n=-cos\u03c03=-cos\u03c03<br \/>\n=cos(\u03c0-\u03c03)=cos(\u03c0-\u03c03)<br \/>\n=cos2\u03c03=cos2\u03c03<br \/>\ni.e., either x=n\u03c03x=n\u03c03 , where n\u2208Zn\u2208Z<br \/>\nor, 2x=2n\u03c0\u00b12\u03c032x=2n\u03c0\u00b12\u03c03 ,where n\u2208Zn\u2208Z .<br \/>\nTherefore, the general solution is n\u03c03n\u03c03 or n\u03c0\u00b1\u03c03,n\u2208Zn\u03c0\u00b1\u03c03,n\u2208Z.<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Class 11 Maths NCERT book solutions for Chapter 3 &#8211; Trigonometric Functions Questions and Answers<\/p>\n","protected":false},"author":21830,"featured_media":121167,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-121198","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121198","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=121198"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121198\/revisions"}],"predecessor-version":[{"id":121216,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121198\/revisions\/121216"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/121167"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=121198"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=121198"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=121198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}