Prove the following identity: 1 - cos q 1 + cos q = cos e c q - c o t q

Given: 1 - cos q 1 + cos q = cos e c q - c o t q L H S = 1 - cos q 1 + cos q Multiply numerator and denominator by 1 - cos q , = 1 - cos q 1 + cos q × 1 - cos q 1 - cos q = 1 - cos q 2 1 + cos q 1 - cos q = 1 - cos q 2 1 - cos 2 q = 1 - cos q 2 sin 2 q = 1 - cos q sin q = 1 sin q - cos q sin q = cos e c q - c o t q = R H S Hence, proved.

Prove the following identity: 1 + cos θ 1 - cos θ - 1 - cos θ 1 + cos θ = 4 c o t θ · cos e c θ

Given: 1 + cos θ 1 - cos θ - 1 - cos θ 1 + cos θ = 4 c o t θ · cos e c θ L H S = 1 + cos θ 1 - cos θ - 1 - cos θ 1 + cos θ = 1 + cos θ 2 - 1 - cos θ 2 1 - cos θ 1 + cos θ = 1 + cos 2 θ + 2 cos θ - ( 1 + cos 2 θ - 2 cos θ ) 1 - cos 2 θ = 1 + cos 2 θ + 2 cos θ - 1 - cos 2 θ + 2 cos θ sin 2 θ = 4 cos θ sin 2 θ = 4 × cos θ sin θ × 1 sin θ = 4 c o t θ · cos e c θ = R H S Hence, proved.

Prove the following identity: sin θ 1 + cos θ + 1 + cos θ sin θ = 2 cos e c θ

Given: sin θ 1 + cos θ + 1 + cos θ sin θ = 2 cos e c θ L H S = sin θ 1 + cos θ + 1 + cos θ sin θ = sin 2 θ + 1 + cos θ 2 1 + cos θ sin θ = sin 2 θ + 1 + cos 2 θ + 2 cos θ 1 + cos θ sin θ = sin 2 θ + cos 2 θ + 1 + 2 cos θ 1 + cos θ sin θ = 1 + 1 + 2 cos θ 1 + cos θ sin θ = 2 + 2 cos θ 1 + cos θ sin θ = 2 1 + cos θ 1 + cos θ sin θ = 2 sin θ = 2 cos e c θ = R H S Hence, proved.

If cos q - sin q = 2 sin q then prove that cos q + sin q = 2 cos q .

cos q - sin q = 2 sin q Squaring on both sides, ( cos q - sin q ) 2 = ( 2 sin q ) 2 cos 2 q + sin 2 q - 2 cos q sin q = 2 sin 2 q 1 - 2 cos q sin q = 2 sin 2 q 2 cos q sin q = 1 - 2 sin 2 q ........ 1 Now, cos q + sin q = 2 cos q Squaring on both sides, ( cos q + sin q ) 2 = ( 2 cos q ) 2 cos 2 q + sin 2 q + 2 cos q sin q = 2 cos 2 q 1 + 2 cos q sin q = 2 cos 2 q 1 + 1 - 2 sin 2 q = 2 cos 2 q From 1 2 - 2 sin 2 q = 2 cos 2 q 2 1 - sin 2 q = 2 cos 2 q ∵ sin 2 q + cos 2 q = 1 2 cos 2 q = 2 cos 2 q Hence, proved.

If tan q = n tan f and sin q = m sin f then prove that cos 2 q = m 2 - 1 n 2 - 1 .

Given: tan q = n tan f sin q = m sin f To prove: cos 2 q = m 2 - 1 n 2 - 1 tan q = n tan f n tan q = 1 tan f n tan q = c o t f ....... 1 Similarly, sin q = m sin f m sin q = 1 sin f m sin q = cos e c f ........ 2 We know that, cos e c 2 θ - c o t 2 θ = 1 Hence, cos e c 2 f - c o t 2 f = 1 ....... 3 Putting values of 1 and 2 in equation 3 , m sin q 2 - n tan q 2 = 1 m sin q 2 - n sin q cos q 2 = 1 m 2 sin 2 q - n 2 cos 2 q sin 2 q = 1 m 2 - n 2 cos 2 q = sin 2 q m 2 - n 2 cos 2 q = 1 - cos 2 q ∵ sin 2 θ + cos 2 θ = 1 m 2 - 1 = n 2 cos 2 q - cos 2 q m 2 - 1 = cos 2 q n 2 - 1 cos 2 q = n 2 - 1 m 2 - 1 Hence, proved.

If x = a cosec q and y = b cot q then prove that x 2 a 2 - y 2 b 2 = 1 .

Given: x = a cosec q y = b cot q To prove: x 2 a 2 - y 2 b 2 = 1 a cosec q 2 a 2 - b cot q 2 b 2 = 1 a 2 cos e c 2 q a 2 - b 2 c o t 2 q b 2 = 1 cos e c 2 q - c o t 2 q = 1 1 sin 2 q - cos 2 q sin 2 q = 1 1 - cos 2 q sin 2 q = 1 sin 2 q sin 2 q = 1 ∵ sin 2 q = 1 - cos 2 q 1 = 1 Hence, proved.

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Mathematics Class - 10>Trigonometry>Chapter 9 - Trigonometric Equation and Identities>Exercise – 1

Chhattisgarh Board Solutions for Exercise 5: Exercise – 1

Author:Chhattisgarh Board

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 5: Exercise – 1

Attempt the free practice questions from Exercise 5: Exercise – 1 with hints and solutions to strengthen your understanding. Mathematics Class - 10 solutions are prepared by Experienced Embibe Experts.