Prove that, cot θ - cos θ cot θ + cos θ = cosec θ - 1 cosec θ + 1 .

We have to prove cot θ - cos θ cot θ + cos θ = cosec θ - 1 cosec θ + 1 . L H S = cot θ - cos θ cot θ + cos θ = cos θ sin θ - cos θ cos θ sin θ + cos θ ∵ cot θ = cos θ sin θ = cos θ 1 sin θ - 1 cos θ 1 sin θ + 1 = 1 sin θ - 1 1 sin θ + 1 = cosec θ - 1 cosec θ + 1 ∵ 1 sin θ = cosec θ = R H S Hence, proved.

Prove that sin θ - cos θ + 1 sin θ + cos θ - 1 = 1 sec θ - tan θ .

We have to prove sin θ - cos θ + 1 sin θ + cos θ - 1 = 1 sec θ - tan θ . L H S = sin θ - cos θ + 1 sin θ + cos θ - 1 On dividing by cos θ in numerator and denominator, we get = sin θ cos θ - 1 + 1 cos θ sin θ cos θ + 1 - 1 cos θ = tan θ - 1 + sec θ tan θ + 1 - sec θ ∵ sin θ cos θ = tan θ and 1 cos θ = sec θ = tan θ + sec θ - 1 tan θ - sec θ + 1 Now, on multiplying by tan θ - sec θ in numerator and denominator, we get = tan θ + sec θ - 1 tan θ - sec θ + 1 × tan 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 2 θ - sec 2 θ - tan θ - sec θ tan θ - sec θ + 1 tan θ - sec θ = - 1 - tan θ - sec θ tan θ - sec θ + 1 tan θ - sec θ ∵ sec 2 θ - tan 2 θ = 1 = - 1 1 + tan θ - sec θ tan θ - sec θ + 1 tan θ - sec θ = - 1 tan θ - sec θ = 1 sec θ - tan θ = R H S Hence, proved.

Prove that cosec A - sin A sec A - cos A = 1 tan A + cot A .

We have to prove cosec A - sin A sec A - cos A = 1 tan A + cot A . L H S = cosec A - sin A sec A - cos A = 1 sin A - sin A 1 cos A - cos A = 1 - sin 2 A sin A 1 - cos 2 A cos A = cos 2 A · sin 2 A sin A · cos A ∵ sin 2 A + cos 2 A = 1 = 1 sin A · cos A sin 2 A · cos 2 A = 1 1 sin A · cos A = 1 sin 2 A + cos 2 A sin A · cos A ∵ 1 = sin 2 A + cos 2 A = 1 sin A cos A + cos A sin A = 1 tan A + cot A ∵ sin A cos A = tan A and cos A sin A = cot A = R H S Hence, proved.

Prove that 1 + sec A sec A = sin 2 A 1 - cos A .

We have to prove 1 + sec A sec A = sin 2 A 1 - cos A . R H S = sin 2 A 1 - cos A = 1 - cos 2 A 1 - cos A ∵ sin 2 A = 1 - cos 2 A = 1 - cos A 1 + cos A 1 - cos A = 1 + cos A = 1 + 1 sec A ∵ cos A = 1 sec A = 1 + sec A sec A = L H S 1 + sec A sec A = sin 2 A 1 - cos A

Prove that, sec A - 1 sec A + 1 = 1 - cos A 1 + cos A .

We have to prove sec A - 1 sec A + 1 = 1 - cos A 1 + cos A . L H S = sec A - 1 sec A + 1 = 1 cos A - 1 1 cos A + 1 ∵ sec A = 1 cos A = 1 - cos A cos A 1 + cos A cos A = 1 - cos A 1 + cos A = R H S Hence, proved.

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Mathematics Class 10>Chapter 11 - Trigonometry>Optional Exercise

Telangana Board Solutions for Chapter: Trigonometry, Exercise 9: Optional Exercise

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Trigonometry, Exercise 9: Optional Exercise

Attempt the practice questions on Chapter 11: Trigonometry, Exercise 9: Optional Exercise with hints and solutions to strengthen your understanding. Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.