If tan 2 A = cot A - 18 ° , where 2 A is an acute angle, find the value of A

Given that, 2 A is an acute angle. ⇒ 2 A < 90 ° So, using trigonometric ratio of complementary angles, we can say that cot 90 ° - 2 A = tan 2 A ∵ cot 90 ° - θ = tan θ Given that, tan 2 A = cot A - 18 ° Now, replace tan 2 A by cot 90 ° - 2 A in the given question. ⇒ cot 90 ° - 2 A = cot A - 18 ° Now, we can compare the degrees from above, we get 90 ° - 2 A = A - 18 ° ⇒ 2 A + A = 90 ° + 18 ° ⇒ 3 A = 108 ° ⇒ A = 108 3 = 36 ° Thus, the value of A is 36 ∘

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

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

If 4 tanθ = 3 , evaluate 4 sinθ - cosθ + 1 4 sinθ + cosθ - 1 .

Given that, tan θ = 3 4 ∴ tan 2 θ = 9 16 We know that, sec 2 θ = 1 + tan 2 θ . ∴ sec 2 θ = 1 + 9 16 = 25 16 ∴ cos 2 θ = 16 25 ∴ cos θ = 4 5 We know that, sin 2 θ = 1 - cos 2 θ ∴ sin 2 θ = 1 - 16 25 = 9 25 ⇒ sin θ = 3 5 Now, 4 sin θ - cos θ + 1 4 sin θ + cos θ - 1 = 4 × 3 5 - 4 5 + 1 4 × 3 5 + 4 5 - 1 = 12 - 4 + 5 12 + 4 - 5 = 13 11 .

Prove that: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A .

To prove: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A Taking L . H . S . sin A - 2 sin 3 A 2 cos 3 A - cos A = sin A 1 - 2 sin 2 A cos A 2 cos 2 A - 1 = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A + cos 2 A [Since sin 2 A + cos 2 A = 1 ] = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A - cos 2 A = tan A cos 2 A - sin 2 A cos 2 A - sin 2 A = tan A = R . H . S . . [proved]

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Introduction to Trigonometry>Exercise

Embibe Experts Solutions for Chapter: Introduction to Trigonometry, Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Introduction to Trigonometry, Exercise 1: Exercise

Attempt the free practice questions on Chapter 8: Introduction to Trigonometry, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.