{"id":119412,"date":"2022-05-04T15:24:30","date_gmt":"2022-05-04T09:54:30","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119412"},"modified":"2022-05-04T15:24:30","modified_gmt":"2022-05-04T09:54:30","slug":"chapter-8-introduction-to-trigonometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-8-introduction-to-trigonometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 8 – Introduction to Trigonometry Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 8.1<\/h2>\n

1. In \u2206 ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
\n(i) sin A, cos A
\n(ii) sin C, cos C<\/h2>\n

Solution:
\nIn a given triangle ABC, right angled at B = \u2220B = 90\u00b0
\nGiven: AB = 24 cm and BC = 7 cm
\nAccording to the Pythagoras Theorem,
\nIn a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.
\nBy applying Pythagoras theorem, we get
\nAC2=AB2+BC2
\nAC2\u00a0= (24)2+72
\nAC2\u00a0= (576+49)
\nAC2\u00a0=\u00a0625cm2
\nAC = \u221a625 = 25
\nTherefore, AC = 25 cm<\/h3>\n

(i) To find Sin (A), Cos (A)
\nWe know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes
\nSin (A) = Opposite side \/Hypotenuse = BC\/AC = 7\/25
\nCosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,
\nCos (A) = Adjacent side\/Hypotenuse = AB\/AC = 24\/25<\/h3>\n

(ii) To find Sin (C), Cos (C)
\nSin (C) = AB\/AC = 24\/25
\nCos (C) = BC\/AC = 7\/25<\/h3>\n

2. In Fig. 8.13, find tan P \u2013 cot R<\/h2>\n

\"\"
\nSolution:
\nIn the given triangle PQR, the given triangle is right angled at Q and the given measures are:
\nPR = 13cm,
\nPQ = 12cm
\nSince the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem
\nAccording to Pythagorean theorem,
\nIn a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.
\nPR2\u00a0= QR2\u00a0+ PQ2
\nSubstitute the values of PR and PQ
\n132\u00a0= QR2+122
\n169 = QR2+144
\nTherefore, QR2\u00a0= 169\u2212144
\nQR2\u00a0= 25
\nQR = \u221a25 = 5
\nTherefore, the side QR = 5 cm
\nTo find tan P \u2013 cot R:
\nAccording to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes
\ntan (P)\u00a0=\u00a0Opposite side \/Adjacent side = QR\/PQ = 5\/12
\nSince cot function is the reciprocal of the tan function, the ratio of cot function becomes,
\nCot (R) = Adjacent side\/Opposite side = QR\/PQ = 5\/12
\nTherefore,
\ntan (P) \u2013 cot (R) = 5\/12 \u2013 5\/12 = 0
\nTherefore, tan(P) \u2013 cot(R) = 0<\/h3>\n

3. If sin A = 3\/4, Calculate cos A and tan A.<\/h2>\n

Solution:
\nLet us assume a right angled triangle ABC, right angled at B
\nGiven: Sin A = 3\/4
\nWe know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.
\nTherefore, Sin A = Opposite side \/Hypotenuse= 3\/4
\nLet BC be 3k and AC will be 4k
\nwhere k is a positive real number.
\nAccording to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,
\nAC2=AB2\u00a0+ BC2
\nSubstitute the value of AC and BC
\n(4k)2=AB2\u00a0+ (3k)2
\n16k2\u22129k2\u00a0=AB2
\nAB2=7k2
\nTherefore, AB = \u221a7k
\nNow, we have to find the value of cos A and tan A
\nWe know that,
\nCos (A) = Adjacent side\/Hypotenuse
\nSubstitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get
\nAB\/AC = \u221a7k\/4k = \u221a7\/4
\nTherefore, cos (A) = \u221a7\/4
\ntan(A) = Opposite side\/Adjacent side
\nSubstitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,
\nBC\/AB = 3k\/\u221a7k = 3\/\u221a7
\nTherefore, tan A = 3\/\u221a7<\/h3>\n

4. Given 15 cot A = 8, find sin A and sec A.<\/h2>\n

Solution:
\nLet us assume a right angled triangle ABC, right angled at B
\nGiven: 15 cot A = 8
\nSo, Cot A = 8\/15
\nWe know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.
\nTherefore, cot A = Adjacent side\/Opposite side = AB\/BC = 8\/15
\nLet AB be 8k and BC will be 15k
\nWhere, k is a positive real number.
\nAccording to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,
\nAC2=AB2\u00a0+ BC2
\nSubstitute the value of AB and BC
\nAC2= (8k)2\u00a0+ (15k)2
\nAC2= 64k2\u00a0+ 225k2
\nAC2= 289k2
\nTherefore, AC = 17k
\nNow, we have to find the value of sin A and sec A
\nWe know that,
\nSin (A) = Opposite side \/Hypotenuse
\nSubstitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get
\nSin A = BC\/AC = 15k\/17k = 15\/17
\nTherefore, sin A = 15\/17
\nSince secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.
\nSec (A) = Hypotenuse\/Adjacent side
\nSubstitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,
\nAC\/AB = 17k\/8k = 17\/8
\nTherefore sec (A) = 17\/8<\/h3>\n

5. Given sec \u03b8 = 13\/12 Calculate all other trigonometric ratios<\/h2>\n

Solution:
\nWe know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side
\nLet us assume a right angled triangle ABC, right angled at B
\nsec \u03b8 =13\/12 = Hypotenuse\/Adjacent side = AC\/AB
\nLet AC be 13k and AB will be 12k
\nWhere, k is a positive real number.
\nAccording to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,
\nAC2=AB2\u00a0+ BC2
\nSubstitute the value of AB and AC
\n(13k)2= (12k)2\u00a0+ BC2
\n169k2= 144k2\u00a0+ BC2
\n169k2= 144k2\u00a0+ BC2
\nBC2 =\u00a0169k2\u00a0\u2013 144k2
\nBC2= 25k2
\nTherefore, BC = 5k
\nNow, substitute the corresponding values in all other trigonometric ratios
\nSo,
\nSin \u03b8 = Opposite Side\/Hypotenuse = BC\/AC = 5\/13
\nCos \u03b8 = Adjacent Side\/Hypotenuse = AB\/AC = 12\/13
\ntan \u03b8 = Opposite Side\/Adjacent Side = BC\/AB = 5\/12
\nCosec \u03b8 = Hypotenuse\/Opposite Side = AC\/BC = 13\/5
\ncot \u03b8 = Adjacent Side\/Opposite Side = AB\/BC = 12\/5<\/h3>\n

6. If \u2220A and \u2220B are acute angles such that cos A = cos B, then show that \u2220 A = \u2220 B.<\/h2>\n

Solution:
\nLet us assume the triangle ABC\u00a0in which\u00a0CD\u22a5AB
\nGive that the angles A and B are acute angles, such that
\nCos (A) = cos (B)
\nAs per the angles taken, the cos ratio is written as
\nAD\/AC = BD\/BC
\nNow, interchange the terms, we get
\nAD\/BD = AC\/BC
\nLet take a constant value
\nAD\/BD = AC\/BC = k
\nNow consider the equation as
\nAD = k BD \u2026(1)
\nAC = k BC \u2026(2)
\nBy applying Pythagoras theorem in\u00a0\u25b3CAD\u00a0and\u00a0\u25b3CBD\u00a0we get,
\nCD2\u00a0= BC2\u00a0\u2013 BD2\u00a0\u2026 (3)
\nCD2\u00a0=AC2\u00a0\u2212AD2\u00a0\u2026.(4)
\nFrom the equations (3) and (4) we get,
\nAC2\u2212AD2\u00a0= BC2\u2212BD2
\nNow substitute the equations (1) and (2) in (3) and (4)
\nK2(BC2\u2212BD2)=(BC2\u2212BD2)\u00a0k2=1
\nPutting this value in equation, we obtain
\nAC = BC
\n\u2220A=\u2220B\u00a0(Angles opposite to equal side are equal-isosceles triangle)<\/h3>\n

7. If cot \u03b8 = 7\/8, evaluate :
\n(i) (1 + sin \u03b8)(1 \u2013 sin \u03b8)\/(1+cos \u03b8)(1-cos \u03b8)
\n(ii) cot2\u00a0\u03b8<\/h2>\n

Solution:
\nLet\u00a0us assume a \u25b3ABC\u00a0in which\u00a0\u2220B = 90\u00b0 and \u2220C = \u03b8
\nGiven:
\ncot \u03b8 = BC\/AB = 7\/8
\nLet BC = 7k and AB = 8k, where k is a positive real number
\nAccording to Pythagoras theorem in\u00a0\u25b3ABC\u00a0we get.
\nAC2\u00a0= AB2+BC2
\nAC2\u00a0= (8k)2+(7k)2
\nAC2\u00a0= 64k2+49k2
\nAC2\u00a0= 113k2
\nAC = \u221a113 k
\nAccording to the sine and cos function ratios, it is written as
\nsin \u03b8 = AB\/AC = Opposite Side\/Hypotenuse = 8k\/\u221a113 k = 8\/\u221a113 and
\ncos \u03b8 = Adjacent Side\/Hypotenuse = BC\/AC = 7k\/\u221a113 k = 7\/\u221a113
\nNow apply the values of sin function and cos function:<\/h3>\n

\"\"
\n8. If 3 cot A = 4, check whether (1-tan2\u00a0A)\/(1+tan2\u00a0A) = cos2\u00a0A \u2013 sin\u00a02\u00a0A or not.<\/h2>\n

Solution:
\nLet\u00a0\u25b3ABC\u00a0in which\u00a0\u2220B=90\u00b0
\nWe know that, cot function is the reciprocal of tan function and it is written as
\ncot(A) = AB\/BC = 4\/3
\nLet AB = 4k an BC =3k, where k is a positive real number.
\nAccording to the Pythagorean theorem,
\nAC2=AB2+BC2
\nAC2=(4k)2+(3k)2
\nAC2=16k2+9k2
\nAC2=25k2
\nAC=5k
\nNow, apply the values corresponding to the ratios
\ntan(A) = BC\/AB = 3\/4
\nsin (A) = BC\/AC = 3\/5
\ncos (A) = AB\/AC = 4\/5
\nNow compare the left hand side(LHS) with right hand side(RHS)<\/h3>\n

\"\"
\nSince, both the LHS and RHS = 7\/25
\nR.H.S. =L.H.S.
\nHence,\u00a0(1-tan2\u00a0A)\/(1+tan2\u00a0A) = cos2\u00a0A \u2013 sin\u00a02\u00a0A\u00a0 is proved<\/h3>\n

9. In triangle ABC, right-angled at B, if tan A = 1\/\u221a3 find the value of:
\n(i) sin A cos C + cos A sin C
\n(ii) cos A cos C \u2013 sin A sin C<\/h2>\n

Solution:
\nLet \u0394ABC in which\u00a0\u2220B=90\u00b0
\ntan A = BC\/AB = 1\/\u221a3
\nLet BC = 1k and AB = \u221a3 k,
\nWhere k is the positive real number of the problem
\nBy Pythagoras theorem in \u0394ABC we get:
\nAC2=AB2+BC2
\nAC2=(\u221a3 k)2+(k)2
\nAC2=3k2+k2
\nAC2=4k2
\nAC = 2k
\nNow find the values of cos A, Sin A
\nSin A = BC\/AC = 1\/2
\nCos A = AB\/AC = \u221a3\/2
\nThen find the values of cos C and sin C
\nSin C = AB\/AC =\u00a0\u221a3\/2
\nCos C = BC\/AC = 1\/2
\nNow, substitute the values in the given problem
\n(i) sin A cos C + cos A sin C = (1\/2) \u00d7(1\/2 )+ \u221a3\/2 \u00d7\u221a3\/2 = 1\/4 + 3\/4 = 1
\n(ii) cos A cos C \u2013 sin A sin C = (\u221a3\/2 )(1\/2) \u2013 (1\/2) (\u221a3\/2 ) = 0<\/h3>\n

10. In \u2206 PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P<\/h2>\n

Solution:
\nIn a given triangle PQR, right angled at Q, the following measures are
\nPQ = 5 cm
\nPR + QR = 25 cm
\nNow let us assume, QR = x
\nPR = 25-QR
\nPR = 25- x
\nAccording to the Pythagorean Theorem,
\nPR2\u00a0= PQ2\u00a0+ QR2
\nSubstitute the value of PR as x
\n(25- x)\u00a02\u00a0= 52\u00a0+ x2
\n252\u00a0+ x2\u00a0\u2013 50x = 25 + x2
\n625 + x2-50x -25 \u2013 x2\u00a0= 0
\n-50x = -600
\nx= -600\/-50
\nx = 12 = QR
\nNow, find the value of PR
\nPR = 25- QR
\nSubstitute the value of QR
\nPR = 25-12
\nPR = 13
\nNow, substitute the value to the given problem
\n(1) sin p = Opposite Side\/Hypotenuse = QR\/PR = 12\/13
\n(2) Cos p = Adjacent Side\/Hypotenuse = PQ\/PR = 5\/13
\n(3) tan p =Opposite Side\/Adjacent side = QR\/PQ = 12\/5<\/h3>\n

11. State whether the following are true or false. Justify your answer.
\n(i) The value of tan A is always less than 1.
\n(ii) sec A = 12\/5 for some value of angle A.
\n(iii)cos A is the abbreviation used for the cosecant of angle A.
\n(iv) cot A is the product of cot and A.
\n(v) sin \u03b8 = 4\/3 for some angle \u03b8.<\/h2>\n

Solution:
\n(i)\u00a0The value of tan A is always less than 1.
\nAnswer:\u00a0False
\nProof: In \u0394MNC in which\u00a0\u2220N\u00a0=\u00a090\u2218,
\nMN = 3, NC = 4 and MC = 5
\nValue of tan M = 4\/3 which is greater than.
\nThe triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.
\nMC2=MN2+NC2
\n52=32+42
\n25=9+16
\n25\u00a0=\u00a025<\/h3>\n

(ii)\u00a0sec A = 12\/5 for some value of angle A
\nAnswer:\u00a0True
\nJustification: Let a \u0394MNC in which \u2220N = 90\u00ba,
\nMC=12k and MB=5k, where k is a positive real number.
\nBy Pythagoras theorem we get,
\nMC2=MN2+NC2
\n(12k)2=(5k)2+NC2
\nNC2+25k2=144k2
\nNC2=119k2
\nSuch a triangle is possible as\u00a0it will follow the Pythagoras theorem.<\/h3>\n

(iii)\u00a0cos A is the abbreviation used for the cosecant of angle A.
\nAnswer:\u00a0False
\nJustification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.<\/h3>\n

(iv)\u00a0cot A is the product of cot and A.
\nAnswer:\u00a0False
\nJustification: cot M is not the product of cot and M. It is the cotangent of\u00a0\u2220M.<\/h3>\n

(v)\u00a0sin \u03b8 = 4\/3 for some angle \u03b8.
\nAnswer: False
\nJustification: sin \u03b8 = Opposite\/Hypotenuse
\nWe know that in a right angled triangle, Hypotenuse is the longest side.
\n\u2234\u00a0sin \u03b8 will always less than 1 and it can never be\u00a04\/3 for any value of\u00a0\u03b8.<\/h3>\n

Exercise 8.2<\/h2>\n

1. Evaluate the following:
\n(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0
\n(ii) 2 tan2\u00a045\u00b0 + cos2\u00a030\u00b0 \u2013 sin2\u00a060<\/h2>\n

\"\"
\nSolution:
\n(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0
\nFirst, find the values of the given trigonometric ratios
\nsin 30\u00b0 = 1\/2
\ncos 30\u00b0 = \u221a3\/2
\nsin 60\u00b0 = 3\/2
\ncos 60\u00b0= 1\/2
\nNow, substitute the values in the given problem
\nsin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0 = \u221a3\/2 \u00d7\u221a3\/2 + (1\/2) \u00d7(1\/2 ) = 3\/4+1\/4 = 4\/4 =1<\/h3>\n

(ii) 2 tan2\u00a045\u00b0 + cos2\u00a030\u00b0 \u2013 sin2\u00a060
\nWe know that, the values of the trigonometric ratios are:
\nsin 60\u00b0 = \u221a3\/2
\ncos 30\u00b0 = \u221a3\/2
\ntan 45\u00b0 = 1
\nSubstitute the values in the given problem
\n2 tan2\u00a045\u00b0 + cos2\u00a030\u00b0 \u2013 sin2\u00a060 = 2(1)2\u00a0+ (\u221a3\/2)2-(\u221a3\/2)2
\n2 tan2\u00a045\u00b0 + cos2\u00a030\u00b0 \u2013 sin2\u00a060 = 2 + 0
\n2 tan2\u00a045\u00b0 + cos2\u00a030\u00b0 \u2013 sin2\u00a060 = 2<\/h3>\n

(iii) cos 45\u00b0\/(sec 30\u00b0+cosec 30\u00b0)
\nWe know that,
\ncos 45\u00b0 = 1\/\u221a2
\nsec 30\u00b0 = 2\/\u221a3
\ncosec 30\u00b0 = 2
\nSubstitute the values, we get<\/h3>\n

\"\"
\nNow, multiply both the numerator and denominator by \u221a2 , we get<\/h3>\n

\"\"
\nTherefore, cos 45\u00b0\/(sec 30\u00b0+cosec 30\u00b0) = (3\u221a2 \u2013 \u221a6)\/8<\/p>\n

\"\"
\nWe know that,
\nsin 30\u00b0 = 1\/2
\ntan 45\u00b0 = 1
\ncosec 60\u00b0 = 2\/\u221a3
\nsec 30\u00b0 = 2\/\u221a3
\ncos 60\u00b0 = 1\/2
\ncot 45\u00b0 = 1
\nSubstitute the values in the given problem, we get<\/h3>\n

\"\"<\/p>\n

\"\"
\nWe know that,
\ncos 60\u00b0 = 1\/2
\nsec 30\u00b0 = 2\/\u221a3
\ntan 45\u00b0 = 1
\nsin 30\u00b0 = 1\/2
\ncos 30\u00b0 = \u221a3\/2
\nNow, substitute the values in the given problem, we get
\n(5cos260\u00b0\u00a0+\u00a04sec230\u00b0\u00a0\u2013 tan245\u00b0)\/(sin2\u00a030\u00b0\u00a0+\u00a0cos2\u00a030\u00b0)
\n= 5(1\/2)2+4(2\/\u221a3)2-12\/(1\/2)2+(\u221a3\/2)2
\n= (5\/4+16\/3-1)\/(1\/4+3\/4)
\n= (15+64-12)\/12\/(4\/4)
\n= 67\/12<\/h3>\n

2.\u00a0Choose the correct option and justify your choice :
\n(i) 2tan 30\u00b0\/1+tan230\u00b0 =
\n(A) sin 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) cos 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) tan 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) sin 30\u00b0<\/h2>\n

(ii) 1-tan245\u00b0\/1+tan245\u00b0 =
\n(A) tan 90\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) sin 45\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) 0<\/h2>\n

(iii)\u00a0\u00a0sin 2A = 2 sin A is true when A =
\n(A) 0\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (B) 30\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 45\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (D) 60\u00b0<\/h2>\n

(iv) 2tan30\u00b0\/1-tan230\u00b0 =
\n(A) cos 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) sin 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (C) tan 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (D) sin 30\u00b0<\/h2>\n

Solution:
\n(i) (A) is correct.
\nSubstitute the of tan 30\u00b0 in the given equation
\ntan 30\u00b0 = 1\/\u221a3
\n2tan 30\u00b0\/1+tan230\u00b0 = 2(1\/\u221a3)\/1+(1\/\u221a3)2
\n= (2\/\u221a3)\/(1+1\/3) = (2\/\u221a3)\/(4\/3)
\n= 6\/4\u221a3\u00a0= \u221a3\/2 = sin 60\u00b0
\nThe obtained solution is equivalent to the trigonometric ratio sin 60\u00b0<\/h3>\n

(ii) (D) is correct.
\nSubstitute the of tan 45\u00b0 in the given equation
\ntan 45\u00b0 = 1
\n1-tan245\u00b0\/1+tan245\u00b0 = (1-12)\/(1+12)
\n= 0\/2 = 0
\nThe solution of the above equation is 0.<\/h3>\n

(iii) (A) is correct.
\nTo find the value of A, substitute the degree given in the options one by one
\nsin 2A = 2 sin A is true when A = 0\u00b0
\nAs sin 2A = sin 0\u00b0 = 0
\n2 sin A = 2 sin 0\u00b0 = 2 \u00d7 0 = 0
\nor,
\nApply the sin 2A formula, to find the degree value
\nsin 2A = 2sin A cos A
\n\u21d22sin A cos A =\u00a02 sin A
\n\u21d2 2cos A = 2 \u21d2 cos A = 1
\nNow, we have to check, to get the solution as 1, which degree value has to be applied.
\nWhen 0 degree is applied to cos value, i.e., cos 0 =1
\nTherefore, \u21d2 A = 0\u00b0<\/h3>\n

(iv) (C) is correct.
\nSubstitute the of tan 30\u00b0 in the given equation
\ntan 30\u00b0 = 1\/\u221a3
\n2tan30\u00b0\/1-tan230\u00b0 = \u00a02(1\/\u221a3)\/1-(1\/\u221a3)2
\n= (2\/\u221a3)\/(1-1\/3) = (2\/\u221a3)\/(2\/3) = \u221a3\u00a0= tan 60\u00b0
\nThe value of the given equation is equivalent to tan 60\u00b0.<\/h3>\n

3. If tan (A + B) = \u221a3 and tan (A \u2013 B) = 1\/\u221a3 ,0\u00b0 < A + B \u2264 90\u00b0; A > B, find A and B.<\/h2>\n

Solution:
\ntan (A + B) = \u221a3
\nSince \u221a3 = tan\u00a060\u00b0
\nNow substitute the degree value
\n\u21d2 tan (A + B) = tan\u00a060\u00b0
\n(A + B) = 60\u00b0 \u2026 (i)
\nThe above equation is assumed as equation (i)
\ntan (A \u2013 B) = 1\/\u221a3
\nSince 1\/\u221a3 = tan\u00a030\u00b0
\nNow substitute the degree value
\n\u21d2 tan (A \u2013 B) = tan 30\u00b0
\n(A \u2013 B) = 30\u00b0 \u2026 equation (ii)
\nNow add the equation (i) and (ii), we get
\nA + B + A \u2013 B = 60\u00b0 + 30\u00b0
\nCancel the terms B
\n2A = 90\u00b0
\nA= 45\u00b0
\nNow, substitute the value of A in equation (i) to find the value of B
\n45\u00b0\u00a0+ B = 60\u00b0
\nB = 60\u00b0 \u2013 45\u00b0
\nB = 15\u00b0
\nTherefore A = 45\u00b0 and B = 15\u00b0<\/h3>\n

4. State whether the following are true or false. Justify your answer.
\n(i) sin (A + B) = sin A + sin B.
\n(ii) The value of sin \u03b8 increases as \u03b8 increases.
\n(iii) The value of cos \u03b8 increases as \u03b8 increases.
\n(iv) sin \u03b8 = cos \u03b8 for all values of \u03b8.
\n(v) cot A is not defined for A = 0\u00b0.<\/h2>\n

Solution:
\n(i) False.
\nJustification:
\nLet us take A = 30\u00b0 and B = 60\u00b0, then
\nSubstitute the values in the sin (A + B) formula, we get
\nsin (A + B) = sin (30\u00b0 + 60\u00b0)\u00a0= sin 90\u00b0\u00a0= 1 and,
\nsin A + sin B = sin 30\u00b0 + sin 60\u00b0
\n= 1\/2\u00a0+ \u221a3\/2 = 1+\u221a3\/2
\nSince the values obtained are not equal, the solution is false.<\/h3>\n

(ii) True.
\nJustification:
\nAccording to the values obtained as per the unit circle, the values of sin are:
\nsin 0\u00b0 = 0
\nsin 30\u00b0 = 1\/2
\nsin 45\u00b0 = 1\/\u221a2
\nsin\u00a060\u00b0 =\u00a0\u221a3\/2
\nsin 90\u00b0 = 1
\nThus the value of sin \u03b8 increases as \u03b8 increases. Hence, the statement is true<\/h3>\n

(iii) False.
\nAccording to the values obtained as per the unit circle, the values of cos are:
\ncos 0\u00b0 = 1
\ncos 30\u00b0 = \u221a3\/2
\ncos 45\u00b0 = 1\/\u221a2
\ncos 60\u00b0 =\u00a01\/2
\ncos 90\u00b0 = 0
\nThus, the value of cos \u03b8 decreases as \u03b8 increases. So, the statement given above is false.<\/h3>\n

(iv) False
\nsin \u03b8 = cos \u03b8,\u00a0when a right triangle has 2 angles of (\u03c0\/4). Therefore, the above statement is false.<\/h3>\n

(v) True.
\nSince cot function is the reciprocal of the tan function, it is also written as:
\ncot A = cos A\/sin A
\nNow substitute A = 0\u00b0
\ncot 0\u00b0 = cos 0\u00b0\/sin 0\u00b0 = 1\/0 = undefined.
\nHence, it is true<\/h3>\n

Exercise 8.3<\/h2>\n

1. Evaluate :
\n(i) sin 18\u00b0\/cos 72\u00b0
\n(ii) tan 26\u00b0\/cot 64\u00b0
\n(iii)\u00a0\u00a0cos 48\u00b0 \u2013 sin 42\u00b0
\n(iv)\u00a0\u00a0cosec 31\u00b0 \u2013 sec 59\u00b0<\/h2>\n

Solution:
\n(i) sin 18\u00b0\/cos 72\u00b0
\nTo simplify this, convert the sin function into cos function
\nWe know that, 18\u00b0 is written as 90\u00b0 \u2013 18\u00b0, which is equal to the cos 72\u00b0.
\n= sin (90\u00b0 \u2013 18\u00b0)\u00a0\/cos 72\u00b0
\nSubstitute the value, to simplify this equation
\n=\u00a0cos 72\u00b0\u00a0\/cos 72\u00b0 = 1<\/h3>\n

(ii) tan 26\u00b0\/cot 64\u00b0
\nTo simplify this, convert the tan function into cot function
\nWe know that, 26\u00b0 is written as 90\u00b0 \u2013 26\u00b0, which is equal to the cot 64\u00b0.
\n= tan (90\u00b0 \u2013 26\u00b0)\/cot 64\u00b0
\nSubstitute the value, to simplify this equation
\n= cot 64\u00b0\/cot 64\u00b0 = 1<\/h3>\n

(iii)\u00a0cos 48\u00b0 \u2013 sin 42\u00b0
\nTo simplify this, convert the cos function into sin function
\nWe know that, 48\u00b0 is written as 90\u00b0 \u2013 42\u00b0, which is equal to the sin 42\u00b0.
\n= cos (90\u00b0 \u2013 42\u00b0) \u2013 sin 42\u00b0
\nSubstitute the value, to simplify this equation
\n= sin 42\u00b0 \u2013 sin 42\u00b0\u00a0= 0<\/h3>\n

(iv) cosec 31\u00b0 \u2013 sec 59\u00b0
\nTo simplify this, convert the cosec function into sec function
\nWe know that, 31\u00b0 is written as 90\u00b0 \u2013 59\u00b0, which is equal to the sec 59\u00b0
\n= cosec (90\u00b0 \u2013 59\u00b0) \u2013 sec 59\u00b0
\nSubstitute the value, to simplify this equation
\n= sec 59\u00b0 \u2013 sec 59\u00b0\u00a0= 0<\/h3>\n

2. \u00a0Show that:
\n(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0 = 1
\n(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/h3>\n

Solution:
\n(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0
\nSimplify the given problem by converting some of the tan functions to the cot functions
\nWe know that, tan 48\u00b0 = tan (90\u00b0 \u2013 42\u00b0) = cot 42\u00b0
\ntan 23\u00b0 = tan (90\u00b0 \u2013 67\u00b0) = cot 67\u00b0
\n= tan (90\u00b0 \u2013 42\u00b0) tan (90\u00b0 \u2013 67\u00b0) tan 42\u00b0 tan 67\u00b0
\nSubstitute the values
\n= cot 42\u00b0 cot 67\u00b0 tan 42\u00b0 tan 67\u00b0
\n= (cot 42\u00b0 tan 42\u00b0) (cot 67\u00b0 tan 67\u00b0)\u00a0= 1\u00d71\u00a0= 1<\/h3>\n

(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0
\nSimplify the given problem by converting some of the cos functions to the sin functions
\nWe know that,
\ncos 38\u00b0 = cos (90\u00b0 \u2013 52\u00b0) = sin 52\u00b0
\ncos 52\u00b0= cos (90\u00b0-38\u00b0) = sin 38\u00b0
\n= cos (90\u00b0 \u2013 52\u00b0) cos (90\u00b0-38\u00b0) \u2013 sin 38\u00b0 sin 52\u00b0
\nSubstitute the values
\n= sin 52\u00b0 sin 38\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0\u00a0= 0<\/h3>\n

3. If tan 2A = cot (A \u2013 18\u00b0), where 2A is an acute angle, find the value of A.<\/h2>\n

Solution:
\ntan 2A = cot (A- 18\u00b0)
\nWe know that tan 2A = cot (90\u00b0 \u2013 2A)
\nSubstitute the above equation in the given problem
\n\u21d2 cot (90\u00b0 \u2013 2A) = cot (A -18\u00b0)
\nNow, equate the angles,
\n\u21d2 90\u00b0 \u2013 2A = A- 18\u00b0\u00a0\u21d2\u00a0108\u00b0 = 3A
\nA = 108\u00b0 \/ 3
\nTherefore, the value of A = 36\u00b0<\/h3>\n

4.\u00a0\u00a0If tan A = cot B, prove that A + B = 90\u00b0.<\/h2>\n

Solution:
\ntan A = cot B
\nWe know that cot B = tan (90\u00b0 \u2013 B)
\nTo prove A + B = 90\u00b0, substitute the above equation in the given problem
\ntan A = tan (90\u00b0 \u2013 B)
\nA = 90\u00b0 \u2013 B
\nA + B = 90\u00b0
\nHence Proved.<\/h3>\n

5. If sec 4A = cosec (A \u2013 20\u00b0), where 4A is an acute angle, find the value of A.<\/h2>\n

Solution:
\nsec 4A = cosec (A \u2013 20\u00b0)
\nWe know that sec 4A = cosec (90\u00b0 \u2013 4A)
\nTo find the value of A, substitute the above equation in the given problem
\ncosec (90\u00b0 \u2013 4A) = cosec (A \u2013 20\u00b0)
\nNow, equate the angles
\n90\u00b0 \u2013 4A= A- 20\u00b0
\n110\u00b0 = 5A
\nA = 110\u00b0\/ 5 = 22\u00b0
\nTherefore, the value of A = 22\u00b0<\/h3>\n

6. If A, B and C are interior angles of a triangle ABC, then show that
\nsin (B+C\/2) = cos A\/2<\/h2>\n

Solution:
\nWe know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180\u00b0
\nA + B\u00a0+ C = 180\u00b0 \u2026.(1)
\nTo find the value of (B+ C)\/2, simplify the equation (1)
\n\u21d2 B\u00a0+ C = 180\u00b0 \u2013 A
\n\u21d2 (B+C)\/2 = (180\u00b0-A)\/2
\n\u21d2 (B+C)\/2 = (90\u00b0-A\/2)
\nNow, multiply both sides by sin functions, we get
\n\u21d2 sin (B+C)\/2 = sin (90\u00b0-A\/2)
\nSince sin (90\u00b0-A\/2) = cos A\/2, the above equation is equal to
\nsin (B+C)\/2 = cos A\/2
\nHence proved.<\/h3>\n

7.\u00a0Express sin 67\u00b0 + cos 75\u00b0 in terms of trigonometric ratios of angles between 0\u00b0 and 45\u00b0.<\/h2>\n

Solution:
\nGiven:
\nsin 67\u00b0 + cos 75\u00b0
\nIn term of sin as cos function and cos as sin function, it can be written as follows
\nsin 67\u00b0 = sin (90\u00b0 \u2013 23\u00b0)
\ncos 75\u00b0 = cos (90\u00b0 \u2013 15\u00b0)
\nSo, sin 67\u00b0 + cos 75\u00b0 = sin (90\u00b0 \u2013 23\u00b0) + cos (90\u00b0 \u2013 15\u00b0)
\nNow, simplify the above equation
\n= cos 23\u00b0 + sin 15\u00b0
\nTherefore, sin 67\u00b0 + cos 75\u00b0 is also expressed as cos 23\u00b0 + sin 15\u00b0<\/h3>\n

Exercise 8.4<\/h2>\n

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.<\/h2>\n

Solution:
\nTo convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas
\nWe know that,
\ncosec2A\u00a0\u2013 cot2A = 1
\ncosec2A = 1\u00a0+ cot2A
\nSince cosec function is the inverse of sin function, it is written as
\n1\/sin2A = 1\u00a0+ cot2A
\nNow, rearrange the terms, it becomes
\nsin2A = 1\/(1+cot2A)
\nNow, take square roots on both sides, we get
\nsin A = \u00b11\/(\u221a(1+cot2A)
\nThe above equation defines the sin function in terms of cot function
\nNow, to express sec function in terms of cot function, use this formula
\nsin2A = 1\/ (1+cot2A)
\nNow, represent the sin function as cos function
\n1 \u2013 cos2A = 1\/ (1+cot2A)
\nRearrange the terms,
\ncos2A = 1 \u2013 1\/(1+cot2A)
\n\u21d2cos2A =\u00a0(1-1+cot2A)\/(1+cot2A)
\nSince sec function is the inverse of cos function,
\n\u21d2 1\/sec2A = cot2A\/(1+cot2A)
\nTake the reciprocal and square roots on both sides, we get
\n\u21d2 sec A = \u00b1\u221a (1+cot2A)\/cotA
\nNow, to express tan function in terms of cot function
\ntan A = sin A\/cos A and cot A = cos A\/sin A
\nSince cot function is the inverse of tan function, it is rewritten as
\ntan A = 1\/cot A<\/h3>\n

2.\u00a0Write all the other trigonometric ratios of \u2220A in terms of sec A.<\/h2>\n

Solution:
\nCos A function in terms of sec A:
\nsec A = 1\/cos A
\n\u21d2 cos A = 1\/sec A
\nsec A function in terms of sec A:
\ncos2A\u00a0+ sin2A = 1
\nRearrange the terms
\nsin2A = 1 \u2013 cos2A
\nsin2A = 1 \u2013 (1\/sec2A)
\nsin2A = (sec2A-1)\/sec2A
\nsin A = \u00b1 \u221a(sec2A-1)\/sec A
\ncosec A function in terms of sec A:
\nsin A = 1\/cosec A
\n\u21d2cosec A = 1\/sin A
\ncosec A = \u00b1 sec A\/\u221a(sec2A-1)
\nNow, tan A function in terms of sec A:
\nsec2A \u2013 tan2A = 1
\nRearrange the terms
\n\u21d2 tan2A = sec2A \u2013 1
\ntan A = \u221a(sec2A\u00a0\u2013 1)
\ncot A function in terms of sec A:
\ntan A = 1\/cot A
\n\u21d2 cot A = 1\/tan A
\ncot A = \u00b11\/\u221a(sec2A\u00a0\u2013 1)<\/h3>\n

3.\u00a0Evaluate:
\n(i) (sin263\u00b0 + sin227\u00b0)\/(cos217\u00b0 + cos273\u00b0)
\n(ii)\u00a0\u00a0sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0<\/h2>\n

Solution:
\n(i) (sin263\u00b0 + sin227\u00b0)\/(cos217\u00b0 + cos273\u00b0)
\nTo simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
\n= [sin2(90\u00b0-27\u00b0) + sin227\u00b0] \/ [cos2(90\u00b0-73\u00b0) + cos273\u00b0)]
\n= (cos227\u00b0\u00a0+ sin227\u00b0)\/(sin227\u00b0\u00a0+ cos273\u00b0)
\n= 1\/1 =1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (since sin2A\u00a0+ cos2A = 1)
\nTherefore, (sin263\u00b0 + sin227\u00b0)\/(cos217\u00b0 + cos273\u00b0) = 1<\/h3>\n

(ii) sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0
\nTo simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
\n= sin(90\u00b0-25\u00b0) cos 65\u00b0\u00a0+ cos (90\u00b0-65\u00b0) sin 65\u00b0
\n= cos 65\u00b0 cos 65\u00b0\u00a0+ sin 65\u00b0 sin 65\u00b0
\n= cos265\u00b0\u00a0+ sin265\u00b0 = 1 (since sin2A\u00a0+ cos2A = 1)
\nTherefore, sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0 = 1<\/h3>\n

4. Choose the correct option. Justify your choice.
\n(i) 9 sec2A \u2013 9 tan2A =
\n(A) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(B) 9 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) 0
\n(ii) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8)
\n(A) 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (B) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) \u2013 1
\n(iii) (sec A + tan A) (1 \u2013 sin A) =
\n(A) sec A \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(B) sin A \u00a0 \u00a0 \u00a0 \u00a0(C) cosec A \u00a0 \u00a0 \u00a0(D) cos A
\n(iv) 1+tan2A\/1+cot2A =
\n(A)\u00a0sec2\u00a0A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) -1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C)\u00a0cot2A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(D) tan2A<\/h2>\n

Solution:
\n(i) (B) is correct.
\nJustification:
\nTake 9 outside, and it becomes
\n9\u00a0sec2A\u00a0\u2013 9 tan2A
\n= 9 (sec2A\u00a0\u2013 tan2A)
\n=\u00a09\u00d71 =\u00a09 \u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\u2235\u00a0sec2 A \u2013 tan2 A = 1)
\nTherefore, 9\u00a0sec2A\u00a0\u2013 9 tan2A = 9<\/h3>\n

(ii) (C) is correct
\nJustification:
\n(1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8)
\nWe know that, tan \u03b8 = sin \u03b8\/cos \u03b8
\nsec \u03b8 = 1\/ cos \u03b8
\ncot \u03b8 = cos \u03b8\/sin \u03b8
\ncosec \u03b8 = 1\/sin \u03b8
\nNow, substitute the above values in the given problem, we get
\n= (1\u00a0+ sin \u03b8\/cos \u03b8\u00a0+ 1\/ cos \u03b8) (1\u00a0+ cos \u03b8\/sin \u03b8 \u2013 1\/sin \u03b8)
\nSimplify the above equation,
\n= (cos \u03b8 +sin \u03b8+1)\/cos \u03b8\u00a0\u00d7 (sin \u03b8+cos \u03b8-1)\/sin \u03b8
\n= (cos \u03b8+sin \u03b8)2-12\/(cos \u03b8 sin \u03b8)
\n= (cos2\u03b8 + sin2\u03b8 + 2cos \u03b8 sin \u03b8 -1)\/(cos \u03b8 sin \u03b8)
\n= (1+ 2cos \u03b8 sin \u03b8 -1)\/(cos \u03b8 sin \u03b8) (Since cos2\u03b8 + sin2\u03b8 = 1)
\n= (2cos \u03b8 sin \u03b8)\/(cos \u03b8 sin \u03b8) = 2
\nTherefore, (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8) =2<\/h3>\n

(iii) (D) is correct.
\nJustification:
\nWe know that,
\nSec A= 1\/cos A
\nTan A = sin A \/ cos A
\nNow, substitute the above values in the given problem, we get
\n(secA + tanA) (1 \u2013 sinA)
\n= (1\/cos A\u00a0+ sin A\/cos A) (1 \u2013 sinA)
\n= (1+sin A\/cos A) (1 \u2013 sinA)
\n= (1 \u2013 sin2A)\/cos A
\n= cos2A\/cos A = cos A
\nTherefore, (secA + tanA) (1 \u2013 sinA) = cos A<\/h3>\n

(iv)\u00a0(D) is correct.
\nJustification:
\nWe know that,
\ntan2A =1\/cot2A
\nNow, substitute this in the given problem, we get
\n1+tan2A\/1+cot2A
\n= (1+1\/cot2A)\/1+cot2A
\n= (cot2A+1\/cot2A)\u00d7(1\/1+cot2A)
\n=\u00a01\/cot2A = tan2A
\nSo, 1+tan2A\/1+cot2A = tan2A<\/h3>\n

5.\u00a0Prove the following identities, where the angles involved are acute angles for which the
\nexpressions are defined.
\n(i) (cosec \u03b8 \u2013 cot \u03b8)2\u00a0= (1-cos \u03b8)\/(1+cos \u03b8)
\n(ii) cos A\/(1+sin A)\u00a0+ (1+sin A)\/cos A = 2 sec A
\n(iii) tan \u03b8\/(1-cot \u03b8)\u00a0+ cot \u03b8\/(1-tan \u03b8) = 1\u00a0+ sec \u03b8 cosec \u03b8
\n[Hint : Write the expression in terms of sin \u03b8 and cos \u03b8]
\n(iv) (1\u00a0+ sec A)\/sec A = sin2A\/(1-cos A)
\n[Hint : Simplify LHS and RHS separately]
\n(v) ( cos A\u2013sin A+1)\/( cos A +sin A\u20131) =\u00a0cosec A + cot A, using the identity cosec2A = 1+cot2A.<\/h2>\n

\"\"
\n(vii) (sin \u03b8 \u2013 2sin3\u03b8)\/(2cos3\u03b8-cos \u03b8) = tan \u03b8
\n(viii) (sin A\u00a0+ cosec A)2\u00a0+ (cos A + sec A)2\u00a0=\u00a07+tan2A+cot2A
\n(ix)\u00a0(cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)
\n[Hint : Simplify LHS and RHS separately]
\n(x) (1+tan2A\/1+cot2A) = (1-tan A\/1-cot A)2\u00a0=\u00a0tan2A<\/h2>\n

Solution:
\n(i) (cosec \u03b8 \u2013 cot \u03b8)2\u00a0= (1-cos \u03b8)\/(1+cos \u03b8)
\nTo prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)
\nL.H.S. = (cosec \u03b8 \u2013 cot \u03b8)2
\nThe above equation is in the form of (a-b)2, and expand it
\nSince (a-b)2\u00a0= a2\u00a0+ b2\u00a0\u2013 2ab
\nHere a = cosec \u03b8 and b = cot \u03b8
\n= (cosec2\u03b8 +\u00a0cot2\u03b8 \u2013 2cosec \u03b8 cot \u03b8)
\nNow, apply the corresponding inverse functions and equivalent ratios to simplify
\n= (1\/sin2\u03b8\u00a0+ cos2\u03b8\/sin2\u03b8 \u2013 2cos \u03b8\/sin2\u03b8)
\n= (1 + cos2\u03b8 \u2013 2cos \u03b8)\/(1 \u2013 cos2\u03b8)
\n= (1-cos \u03b8)2\/(1 \u2013 cos\u03b8)(1+cos \u03b8)
\n= (1-cos \u03b8)\/(1+cos \u03b8) = R.H.S.
\nTherefore, (cosec \u03b8 \u2013 cot \u03b8)2\u00a0= (1-cos \u03b8)\/(1+cos \u03b8)
\nHence proved.<\/h3>\n

(ii) \u00a0(cos A\/(1+sin A))\u00a0+ ((1+sin A)\/cos A) = 2 sec A
\nNow, take the L.H.S of the given equation.
\nL.H.S. = (cos A\/(1+sin A))\u00a0+ ((1+sin A)\/cos A)
\n= [cos2A\u00a0+ (1+sin A)2]\/(1+sin A)cos A
\n= (cos2A + sin2A + 1\u00a0+ 2sin A)\/(1+sin A) cos A
\nSince cos2A + sin2A = 1, we can write it as
\n= (1\u00a0+ 1 + 2sin A)\/(1+sin A) cos A
\n= (2+ 2sin A)\/(1+sin A)cos A
\n= 2(1+sin A)\/(1+sin A)cos A
\n= 2\/cos A = 2 sec A = R.H.S.
\nL.H.S. = R.H.S.
\n(cos A\/(1+sin A))\u00a0+ ((1+sin A)\/cos A) = 2 sec A
\nHence proved.<\/h3>\n

(iii) tan \u03b8\/(1-cot \u03b8)\u00a0+ cot \u03b8\/(1-tan \u03b8) = 1\u00a0+ sec \u03b8 cosec \u03b8
\nL.H.S. = tan \u03b8\/(1-cot \u03b8)\u00a0+ cot \u03b8\/(1-tan \u03b8)
\nWe know that tan \u03b8 =sin \u03b8\/cos \u03b8
\ncot \u03b8 = cos \u03b8\/sin \u03b8
\nNow, substitute it in the given equation, to convert it in a simplified form
\n= [(sin \u03b8\/cos \u03b8)\/1-(cos \u03b8\/sin \u03b8)]\u00a0+ [(cos \u03b8\/sin \u03b8)\/1-(sin\u00a0\u03b8\/cos \u03b8)]
\n= [(sin \u03b8\/cos \u03b8)\/(sin \u03b8-cos \u03b8)\/sin \u03b8]\u00a0+ [(cos \u03b8\/sin \u03b8)\/(cos \u03b8-sin\u00a0\u03b8)\/cos \u03b8]
\n= sin2\u03b8\/[cos \u03b8(sin \u03b8-cos \u03b8)] + cos2\u03b8\/[sin \u03b8(cos \u03b8-sin \u03b8)]
\n= sin2\u03b8\/[cos \u03b8(sin \u03b8-cos \u03b8)] \u2013 cos2\u03b8\/[sin \u03b8(sin \u03b8-cos \u03b8)]
\n= 1\/(sin \u03b8-cos \u03b8) [(sin2\u03b8\/cos \u03b8) \u2013 (cos2\u03b8\/sin \u03b8)]
\n= 1\/(sin \u03b8-cos \u03b8)\u00a0\u00d7 [(sin3\u03b8 \u2013 cos3\u03b8)\/sin\u00a0\u03b8 cos \u03b8]
\n= [(sin \u03b8-cos \u03b8)(sin2\u03b8+cos2\u03b8+sin\u00a0\u03b8 cos \u03b8)]\/[(sin \u03b8-cos \u03b8)sin\u00a0\u03b8 cos \u03b8]
\n= (1\u00a0+ sin\u00a0\u03b8 cos \u03b8)\/sin\u00a0\u03b8 cos \u03b8
\n= 1\/sin\u00a0\u03b8 cos \u03b8\u00a0+ 1
\n= 1\u00a0+ sec \u03b8 cosec \u03b8 = R.H.S.
\nTherefore, L.H.S. = R.H.S.
\nHence proved<\/h3>\n

(iv) \u00a0(1\u00a0+ sec A)\/sec A = sin2A\/(1-cos A)
\nFirst find the simplified form of L.H.S
\nL.H.S. = (1\u00a0+ sec A)\/sec A
\nSince secant function is the inverse function of cos function and it is written as
\n= (1\u00a0+ 1\/cos A)\/1\/cos A
\n= (cos A + 1)\/cos A\/1\/cos A
\nTherefore, (1\u00a0+ sec A)\/sec A\u00a0= cos A\u00a0+ 1
\nR.H.S. = sin2A\/(1-cos A)
\nWe know that sin2A = (1 \u2013 cos2A), we get
\n= (1 \u2013 cos2A)\/(1-cos A)
\n= (1-cos A)(1+cos A)\/(1-cos A)
\nTherefore,\u00a0sin2A\/(1-cos A)= cos A\u00a0+ 1
\nL.H.S. = R.H.S.
\nHence proved<\/h3>\n

(v) (cos A\u2013sin A+1)\/(cos A+sin A\u20131) =\u00a0cosec A + cot A, using the identity cosec2A = 1+cot2A.
\nWith the help of identity function, cosec2A = 1+cot2A, let us prove the above equation.
\nL.H.S. = (cos A\u2013sin A+1)\/(cos A+sin A\u20131)
\nDivide the numerator and denominator by sin A, we get
\n= (cos A\u2013sin A+1)\/sin A\/(cos A+sin A\u20131)\/sin A
\nWe know that cos A\/sin A = cot A and 1\/sin A = cosec A
\n= (cot A \u2013 1 +\u00a0cosec A)\/(cot A+ 1 \u2013 cosec A)
\n= (cot A \u2013 cosec2A\u00a0+ cot2A +\u00a0cosec A)\/(cot A+ 1 \u2013 cosec A) (using cosec2A \u2013 cot2A = 1
\n= [(cot A +\u00a0cosec A) \u2013 (cosec2A\u00a0\u2013 cot2A)]\/(cot A+ 1 \u2013 cosec A)
\n= [(cot A +\u00a0cosec A) \u2013\u00a0(cosec A\u00a0+ cot A)(cosec A\u00a0\u2013 cot A)]\/(1 \u2013 cosec A +\u00a0cot A)
\n= \u00a0(cot A\u00a0+\u00a0cosec A)(1 \u2013 cosec A\u00a0+\u00a0cot A)\/(1 \u2013 cosec A\u00a0+\u00a0cot A)
\n= \u00a0cot A\u00a0+\u00a0cosec A = R.H.S.
\nTherefore, (cos A\u2013sin A+1)\/(cos A+sin A\u20131) =\u00a0cosec A + cot A
\nHence Proved<\/h3>\n

\"\"
\nFirst divide the numerator and denominator of L.H.S. by cos A,<\/h3>\n

\"\"
\nWe know that 1\/cos A = sec A and sin A\/ cos A = tan A and it becomes,
\n= \u221a(sec A+ tan A)\/(sec A-tan A)
\nNow using rationalization, we get<\/h3>\n

\"\"
\n= (sec A + tan A)\/1
\n= sec A + tan A = R.H.S
\nHence proved<\/h3>\n

(vii) (sin \u03b8 \u2013 2sin3\u03b8)\/(2cos3\u03b8-cos \u03b8) = tan \u03b8
\nL.H.S. = (sin \u03b8 \u2013 2sin3\u03b8)\/(2cos3\u03b8 \u2013 cos \u03b8)
\nTake sin \u03b8 as in numerator and cos \u03b8 in denominator as outside, it becomes
\n= [sin \u03b8(1 \u2013 2sin2\u03b8)]\/[cos \u03b8(2cos2\u03b8- 1)]
\nWe know that sin2\u03b8 = 1-cos2\u03b8
\n= sin \u03b8[1 \u2013 2(1-cos2\u03b8)]\/[cos \u03b8(2cos2\u03b8 -1)]
\n=\u00a0[sin \u03b8(2cos2\u03b8 -1)]\/[cos \u03b8(2cos2\u03b8 -1)]
\n= tan \u03b8 = R.H.S.
\nHence proved<\/h3>\n

(viii) (sin A\u00a0+ cosec A)2\u00a0+ (cos A + sec A)2\u00a0=\u00a07+tan2A+cot2A
\nL.H.S. = (sin A\u00a0+ cosec A)2\u00a0+ (cos A + sec A)2
\nIt is of the form (a+b)2, expand it
\n(a+b)2\u00a0=a2\u00a0+ b2\u00a0+2ab
\n= (sin2A\u00a0+ cosec2A\u00a0+ 2 sin A cosec A)\u00a0+ (cos2A\u00a0+\u00a0sec2A\u00a0+ 2 cos A sec A)
\n= (sin2A + cos2A) + 2 sin A(1\/sin A)\u00a0+ 2 cos A(1\/cos A)\u00a0+ 1 + tan2A + 1\u00a0+ cot2A
\n= 1\u00a0+ 2\u00a0+ 2\u00a0+ 2 + tan2A\u00a0+\u00a0cot2A
\n= 7+tan2A+cot2A = R.H.S.
\nTherefore, (sin A\u00a0+ cosec A)2\u00a0+ (cos A + sec A)2\u00a0=\u00a07+tan2A+cot2A
\nHence proved.<\/h3>\n

(ix) (cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)
\nFirst, find the simplified form of L.H.S
\nL.H.S. = (cosec A \u2013 sin A)(sec A \u2013 cos A)
\nNow, substitute the inverse and equivalent trigonometric ratio forms
\n= (1\/sin A \u2013 sin A)(1\/cos A \u2013 cos A)
\n= [(1-sin2A)\/sin A][(1-cos2A)\/cos A]
\n= (cos2A\/sin A)\u00d7(sin2A\/cos A)
\n= cos A sin A
\nNow, simplify the R.H.S
\nR.H.S. = 1\/(tan A+cotA)
\n= 1\/(sin A\/cos A\u00a0+cos A\/sin A)
\n= 1\/[(sin2A+cos2A)\/sin A cos A]
\n= cos A sin A
\nL.H.S. = R.H.S.
\n(cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)
\nHence proved<\/h3>\n

(x) \u00a0(1+tan2A\/1+cot2A) = (1-tan A\/1-cot A)2\u00a0=\u00a0tan2A
\nL.H.S. = (1+tan2A\/1+cot2A)
\nSince cot function is the inverse of tan function,
\n= (1+tan2A\/1+1\/tan2A)
\n= 1+tan2A\/[(1+tan2A)\/tan2A]
\nNow cancel the 1+tan2A terms, we get
\n= tan2A
\n(1+tan2A\/1+cot2A) = tan2A
\nSimilarly,
\n(1-tan A\/1-cot A)2\u00a0=\u00a0tan2A
\nHence proved<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 8 – Introduction to Trigonometry Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119412","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119412","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=119412"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119412\/revisions"}],"predecessor-version":[{"id":119434,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119412\/revisions\/119434"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119412"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119412"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}