Prove the following identity. 1 + 1 tan 2 A 1 + 1 cot 2 A = 1 sin 2 A - sin 4 A

LHS: 1 + 1 tan 2 A 1 + 1 cot 2 A ⇒ 1 + 1 sin 2 A cos 2 A 1 + 1 cos 2 A sin 2 A = 1 + cos 2 A sin 2 A 1 + sin 2 A cos 2 A = sin 2 A + cos 2 A sin 2 A cos 2 A + sin 2 A cos 2 A = 1 sin 2 A 1 cos 2 A = 1 sin 2 A 1 1 - sin 2 A = 1 sin 2 A - sin 4 A = RHS Therefore, 1 + 1 tan 2 A 1 + 1 cot 2 A = 1 sin 2 A - sin 4 A

Prove the following identity. sin 3 A + cos 3 A sin A + cos A = 1 - sin A cos A

Given: LHS= sin 3 A + cos 3 A sin A + cos A We know that, a 3 + b 3 = ( a + b ) ( a 2 - a b + b 2 ) So, sin 3 A + cos 3 A = ( sin A + cos A ) ( sin 2 A - sin A cos A + cos 2 A ) So, sin 3 A + cos 3 A sin A + cos A = ( sin A + cos A ) ( sin 2 A - sin A cos A + cos 2 A ) sin A + cos A = ( sin 2 A - sin A cos A + cos 2 A ) Also, we know that, sin 2 θ + cos 2 θ = 1 So, ( sin 2 A - sin A cos A + cos 2 A ) = ( 1 - sin A cos A ) =RHS

Prove the following identity. 1 + cos θ + sin θ 1 + cos θ - sin θ = 1 + sin θ cos θ

Given: LHS: 1 + cos θ + sin θ 1 + cos θ - sin θ Dividing numerator and denominator by cos θ , we get 1 + cos θ + sin θ cos θ 1 + cos θ - sin θ cos θ ⇒ 1 cos θ + cos θ cos θ + sin θ cos θ 1 cos θ + cos θ cos θ - sin θ cos θ = sec θ + 1 + tan θ sec θ + 1 - tan θ We know that, sec 2 θ - tan 2 θ = 1 Substituting the value of 1 in the numerator, we get sec θ + sec 2 θ - tan 2 θ + tan θ sec θ + 1 - tan θ Also, we know that, a 2 - b 2 = ( a + b ) ( a - b ) ⇒ sec 2 θ - tan 2 θ = ( sec θ + tan θ ) ( sec θ - tan θ ) ⇒ sec θ + tan θ + ( sec θ + tan θ ) ( sec θ - tan θ ) sec θ + 1 - tan θ ⇒ ( sec θ + tan θ ) ( 1 + sec θ - tan θ ) sec θ + 1 - tan θ = sec θ + tan θ We know that, sec θ = 1 cos θ and tan θ = sin θ cos θ ⇒ 1 cos θ + sin θ cos θ = 1 + sin θ cos θ =RHS

Prove the following identity. cos A 1 - tan A + sin A 1 - cot A = sin A + cos A

LHS: cos A 1 - tan A + sin A 1 - cot A We know that, tan A = sin A cos A and cot A = cos A sin A Substituting the above values in the given LHS, we get cos A 1 - sin A cos A + sin A 1 - cos A sin A ⇒ cos A cos A - sin A cos A + sin A sin A - cos A sin A ⇒ cos 2 A cos A - sin A + sin 2 A sin A - cos A ⇒ cos 2 A - sin 2 A cos A - sin A ⇒ ( cos A + sin A ) ( cos A - sin A ) cos A - sin A ⇒ cos A + sin A = RHS Therefore, cos A 1 - tan A + sin A 1 - cot A = sin A + cos A

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Mathematics for Class 10>Chapter 10 - Trigonometry>EXERCISE 10.2

Manipur Board Solutions for Chapter: Trigonometry, Exercise 2: EXERCISE 10.2

Author:Manipur Board

Manipur Board Mathematics Solutions for Exercise - Manipur Board Solutions for Chapter: Trigonometry, Exercise 2: EXERCISE 10.2

Attempt the practice questions on Chapter 10: Trigonometry, Exercise 2: EXERCISE 10.2 with hints and solutions to strengthen your understanding. Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.