If cot ⁡ θ = 1 3 , write the value of 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ .

Given: cot ⁡ θ = 1 3 Let us assume that the Base of the triangle = x and Perpendicular = 3 x Applying Pythagoras Theorem, H y p o t e n u s e = P e r p e n d i c u l a r 2 + B a s e 2 ⇒ H y p o t e n u s e = 2 x Now we find, 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ = 1 - 1 2 2 2 2 - 3 2 2 2 = 1 - 1 4 2 - 3 4 = 3 5 Hence, the value of 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ is 3 5 .

If tan ⁡ A = 3 4 and A + B = 90 o , then what is the value of cot ⁡ B ?

Given that: A + B = 90 o tan ⁡ A = 3 4 A + B = 90 o ⇒ B = 90 o - A ⇒ cot ⁡ B = cot ⁡ 90 o - A ⇒ cot ⁡ B = tan ⁡ A ⇒ cot ⁡ B = 3 4 Hence, the value of cot ⁡ B is 3 4 .

If A + B = 90 ° and cos ⁡ B = 3 5 , what is the value of sin ⁡ A ?

We have: A + B = 90 ° cos ⁡ B = 3 5 A + B = 90 ° ⇒ A = 90 ° - B ⇒ sin A = sin 90 ° - B ⇒ sin ⁡ A = cos ⁡ B ⇒ sin ⁡ A = 3 5 Hence, the value of sin ⁡ A is 3 5 .

Write the acute angle θ satisfying 3 sin ⁡ θ = cos ⁡ θ .

We have: 3 sin ⁡ θ = cos ⁡ θ . ⇒ 3 sin ⁡ θ = cos ⁡ θ ⇒ sin ⁡ θ cos ⁡ θ = 1 3 ⇒ tan θ = tan 30 ° ( ∵ tan 30 ° = 1 3 ) ⇒ θ = 30 ° Hence, the acute angle is 30 ° .

If A + B = 90 ° and tan ⁡ A = 3 4 , what is cot ⁡ B ?

Given in question: A + B = 90 ° tan ⁡ A = 3 4 A + B = 90 ° ⇒ B = 90 ° - A ⇒ cot B = cot 90 ° - A ⇒ cot ⁡ B = tan ⁡ A ⇒ cot B = 3 4 cot 90 ° - A = tan A Hence, the value of cot ⁡ B is 3 4 .

If tan ⁡ A = 5 12 , find the value of sin ⁡ A + cos ⁡ A sec ⁡ A .

Given: tan ⁡ A = 5 12 ⇒ Perpendicular Base = 5 12 Let, perpendicular = 5 k , Base = 12 k . Consider the below diagram, H y p o t e n u s e = P e r p e n d i c u l a r 2 + B a s e 2 ⇒ H y p o t e n u s e = 169 k 2 ⇒ H y p o t e n u s e = 13 k Now we find, sin ⁡ A + cos ⁡ A sec ⁡ A ⇒ sin ⁡ A + cos ⁡ A sec ⁡ A = 5 13 + 12 13 × 13 12 ⇒ sin ⁡ A + cos ⁡ A sec ⁡ A = 17 13 × 13 12 ⇒ sin ⁡ A + cos ⁡ A sec ⁡ A = 17 12 Hence, the value of sin ⁡ A + cos ⁡ A sec ⁡ A is 17 12 .

E M B I B E

MATHEMATICS CLASS X>Chapter 10 - Trigonometric Ratios>VERY SHORT ANSWER TYPE QUESTIONS (VSAQs)

R. D. Sharma Solutions for Chapter: Trigonometric Ratios, Exercise 4: VERY SHORT ANSWER TYPE QUESTIONS (VSAQs)

Author:R. D. Sharma

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Trigonometric Ratios, Exercise 4: VERY SHORT ANSWER TYPE QUESTIONS (VSAQs)

Attempt the free practice questions on Chapter 10: Trigonometric Ratios, Exercise 4: VERY SHORT ANSWER TYPE QUESTIONS (VSAQs) with hints and solutions to strengthen your understanding. MATHEMATICS CLASS X solutions are prepared by Experienced Embibe Experts.