Prove that: 1 sec θ - tan θ = sec θ + tan θ

Given: 1 sec θ - tan θ = sec θ + tan θ 1 1 cos θ - sin θ cos θ = sec θ + tan θ 1 1 - sin θ cos θ = sec θ + tan θ cos θ 1 - sin θ = sec θ + tan θ Multiply L H S with 1 + sin θ , = cos θ 1 - sin θ × 1 + sin θ 1 + sin θ = cos θ + cos θ sin θ 1 2 - sin 2 θ ∵ sin 2 θ + cos 2 θ = 1 = cos θ 1 + sin θ cos 2 θ = 1 + sin θ cos θ = 1 cos θ + sin θ cos θ = sec θ + tan θ R . H . S . = L . H . S . Hence, proved.

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

Given: sec θ + tan θ = cos θ 1 - sin θ 1 cos θ + sin θ cos θ = cos θ 1 - sin θ 1 + sin θ cos θ = cos θ 1 - sin θ L H S is multiplied with ( 1 - sin θ ) , 1 + sin θ cos θ × ( 1 - sin θ ) ( 1 - sin θ ) = cos θ 1 - sin θ 1 - sin 2 θ cos θ ( 1 - sin θ ) = cos θ 1 - sin θ ∵ a 2 - b 2 = a + b a - b cos 2 θ cos ( 1 - sin θ ) = cos θ 1 - sin θ ∵ sin 2 θ + cos 2 θ = 1 cos θ 1 - sin θ = cos θ 1 - sin θ 1 = 1 R . H . S = L . H . S Hence, proved.

Prove that: If tan θ + 1 tan θ = 2 , then show that tan 2 θ + 1 tan 2 θ = 2 .

Given: tan θ + 1 tan θ = 2 Squaring on both sides, tan θ + 1 tan θ 2 = 4 By using the ( a + b ) 2 formula, tan 2 θ + 1 tan 2 θ + 2 × tan θ × 1 tan θ = 4 tan 2 θ + 1 tan 2 θ = 4 - 2 tan 2 θ + 1 tan 2 θ = 2 L H S = R H S Hence, proved.

Prove that: tan A 1 + tan 2 A 2 + cot A 1 + cot 2 A 2 = sin A cos A

Given: tan A 1 + tan 2 A 2 + cot A 1 + cot 2 A 2 L H S = tan A sec 2 A 2 + cot A cosec 2 A 2 = tan A 1 cos 2 A 2 + cot A 1 sin 2 A 2 = sin A cos A 1 cos 2 A 2 + cos A sin A 1 sin 2 A 2 = sin A cos A × cos 4 A + cos A sin A × sin 4 A = sin A cos 3 A + cos A sin 3 A = sin A cos A cos 2 A + sin 2 A ∵ sin 2 θ + cos 2 θ = 1 = sin A cos A = R H S Hence, proved.

Prove that: tan θ sec θ - 1 = tan θ + sec θ + 1 tan θ + sec θ - 1

Given: tan θ sec θ - 1 = tan θ + s e c θ + 1 tan θ + s e c θ - 1 Let us Consider, x = tan θ sec θ - 1 Since, we know that 1 + tan 2 θ = sec 2 θ tan 2 θ = sec 2 θ - 1 ∴ tan 2 θ = sec θ + 1 sec θ - 1 tan θ tan θ = sec θ + 1 sec θ - 1 tan θ sec θ - 1 = sec θ + 1 tan θ By theorem of ratio, a b = c d = a + c b + d tan θ sec θ - 1 = sec θ + 1 tan θ = tan θ + sec θ + 1 sec θ - 1 + tan θ Hence, proved.

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Geometry Standard X>Chapter 6 - Trigonometry>Practice set 6.1

Maharashtra Board Solutions for Chapter: Trigonometry, Exercise 1: Practice set 6.1

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Trigonometry, Exercise 1: Practice set 6.1

Attempt the practice questions on Chapter 6: Trigonometry, Exercise 1: Practice set 6.1 with hints and solutions to strengthen your understanding. Geometry Standard X solutions are prepared by Experienced Embibe Experts.