For what values of A in the first quadrant, the expression cot 3 A - 3 cot A 3 cot 2 A - 1 is positive ?

To find values of A for which cot 3 A - 3 cot A 3 cot 2 A - 1 > 0 . We know that cot 3 A - 3 cot A 3 cot 2 A - 1 = cot 3 A . So, cot 3 A > 0 Given that 0 < A < π 2 , in this we have cot A is positive. For cot 3 A to be positive, we need to have 0 < 3 A < π 2 ⇒ 0 < A < π 6 We also know that cot x is positive in third quadrant. So, if we have π < 3 A < 3 π 2 , then cot 3 A is positive. So, π < 3 A < 3 π 2 ⇒ π 3 < A < π 2 So, the required values of A are 0 < A < π 6 or π 3 < A < π 2 .

Prove that 1 cos 290 ° + 1 3 sin 250 0 = 4 3 .

We know that cos 290 ° = cos 270 ° + 20 ° = sin 20 ° sin 250 ° = sin 270 ° - 20 ° = - cos 20 ° Now, 1 cos 290 ° + 1 3 sin 250 0 = 1 sin 20 ° - 1 3 cos 20 ° = 3 cos 20 ° - sin 20 ° 3 sin 20 ° cos 20 ° = 2 3 2 cos 20 ° - 1 2 sin 20 ° 3 2 2 sin 20 ° cos 20 ° = 4 sin 60 ° cos 20 ° - cos 60 ° sin 20 ° 3 2 sin 20 ° cos 20 ° ∵ sin A cos B - cos A sin B = sin A - B & 2 sin A cos A = sin 2 A = 4 3 · sin 40 ° sin 40 ° = 4 3 Hence, proved that 1 cos 290 ° + 1 3 sin 250 0 = 4 3 .

Prove that sin 2 A ( 1 - cos 2 A ) · ( 1 - cos A ) cos A = tan A 2 .

Given, sin 2 A ( 1 - cos 2 A ) · ( 1 - cos A ) cos A = 2 sin A cos A 2 sin 2 A · ( 1 - cos A ) cos A ∵ sin 2 A = 2 sin A cos A & cos 2 A = 1 - 2 sin 2 A = ( 1 - cos A ) sin A = 2 sin 2 A 2 2 sin A 2 cos A 2 = sin A 2 cos A 2 = tan A 2 Hence, proved that sin 2 A ( 1 - cos 2 A ) · ( 1 - cos A ) cos A = tan A 2 .

Prove that sin 2 x ( sec x + 1 ) · sec 2 x ( sec 2 x + 1 ) = tan x 2 .

Given, sin 2 x ( sec x + 1 ) · sec 2 x ( sec 2 x + 1 ) = sin 2 x 1 cos x + 1 × 1 cos 2 x 1 cos 2 x + 1 = sin 2 x 1 + cos x cos x × 1 cos 2 x × cos 2 x 1 + cos 2 x = sin 2 x cos x 1 + cos x × 1 1 + cos 2 x = 2 sin x cos 2 x 1 + cos x × 1 2 cos 2 x ∵ sin 2 x = 2 sin x cos x & cos 2 x = 2 cos 2 x - 1 = sin x 1 + cos x = 2 sin x 2 cos x 2 2 cos 2 x 2 = sin x 2 cos x 2 = tan x 2 Hence, proved that sin 2 x ( sec x + 1 ) · sec 2 x ( sec 2 x + 1 ) = tan x 2 .

Prove that cos 3 θ - cos 3 θ cos θ + sin 3 θ + sin 3 θ sin θ = 3 .

Given, cos 3 θ - cos 3 θ cos θ + sin 3 θ + sin 3 θ sin θ = cos 3 θ - 4 cos 3 θ - 3 cos θ cos θ + sin 3 θ + 3 sin θ - 4 sin 3 θ sin θ = cos 3 θ - 4 cos 3 θ + 3 cos θ cos θ + sin 3 θ + 3 sin θ - 4 sin 3 θ sin θ = - 3 cos 3 θ + 3 cos θ cos θ + 3 sin θ - 3 sin 3 θ sin θ = cos θ - 3 cos 2 θ + 3 cos θ + sin θ 3 - 3 sin 2 θ sin θ = - 3 cos 2 θ + 3 + 3 - 3 sin 2 θ = 6 - 3 cos 2 θ + sin 2 θ = 6 - 3 1 = 3 Hence, proved that cos 3 θ - cos 3 θ cos θ + sin 3 θ + sin 3 θ sin θ = 3 .

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Intermediate First Year Mathematics Paper 1A>Algebra, Vector Algebra, Trigonometry>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (d)

Andhra Pradesh Board Solutions for Exercise 4: Exercise 6 (d)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 4: Exercise 6 (d)

Attempt the free practice questions from Exercise 4: Exercise 6 (d) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1A solutions are prepared by Experienced Embibe Experts.