If tan ⁡ α − tan ⁡ β = m and cot ⁡ α − cot ⁡ β = n , then prove that cot ⁡ ( α − β ) = 1 m − 1 n .

Given: tan ⁡ α − tan ⁡ β = m and cot ⁡ α − cot ⁡ β = n Consider, cot ⁡ α − cot ⁡ β = n ⇒ 1 tan ⁡ α − 1 tan ⁡ β = n ∵ cot x = 1 tan x ⇒ tan ⁡ β − tan ⁡ α tan ⁡ α tan ⁡ β = n ⇒ − m tan ⁡ α tan ⁡ β = n [ ∵ tan ⁡ α − tan ⁡ β = m ] ⇒ tan ⁡ α tan ⁡ β = − m n − ( 1 ) ∴ cot ⁡ ( α − β ) = 1 tan ⁡ ( α − β ) = 1 + tan ⁡ α tan ⁡ β tan ⁡ α − tan ⁡ β ∵ tan ⁡ ( x − y ) = tan ⁡ x − tan ⁡ y 1 + tan ⁡ x tan ⁡ y = 1 − m n m using 1 = n − m m n = 1 m − 1 n Hence proved.

If tan ⁡ ( α − β ) = 7 24 and tan ⁡ α = 4 3 , where α and β are in the first quadrant prove that α + β = π 2 .

Given: tan ⁡ ( α − β ) = 7 24 and tan ⁡ α = 4 3 , 0 < α , β < π 2 Consider, tan ⁡ ( α − β ) = 7 24 ⇒ tan ⁡ α − tan ⁡ β 1 + tan ⁡ α tan ⁡ β = 7 24 ∵ tan ⁡ ( x − y ) = tan ⁡ x − tan ⁡ y 1 + tan ⁡ x tan ⁡ y ⇒ 4 3 − tan ⁡ β 1 + 4 3 tan ⁡ β = 7 24 ∵ tan ⁡ α = 4 3 ⇒ 4 3 − tan ⁡ β = 7 24 1 + 4 3 tan ⁡ β ⇒ 4 3 − tan ⁡ β = 7 24 + 7 18 tan ⁡ β ⇒ 4 3 − 7 24 = 7 18 tan ⁡ β + tan ⁡ β ⇒ 32 − 7 24 = tan ⁡ β 7 18 + 1 ⇒ 25 24 = 25 18 tan ⁡ β ⇒ tan ⁡ β = 3 4 Now, tan ⁡ α = 4 3 = Perpendicular ( P ) Base ( B ) ∴ Hypotenuse = P 2 + B 2 = 4 2 + 3 2 = 25 = 5 and tan ⁡ β = 3 4 = P 1 B 1 ∴ H 1 = P 1 2 + B 2 2 = 3 2 + 4 2 = 25 = 5 Hence, sin ⁡ α = P H = 4 5 and cos ⁡ α = B H = 3 5 and sin ⁡ β = P 1 H 1 = 3 5 and cos ⁡ β = B 1 H 1 = 4 5 Consider, sin ⁡ ( α + B ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β ∵ sin x + y = sin x cos y + cos x sin y ⇒ sin ⁡ ( α + β ) = 4 5 × 4 5 + 3 5 × 3 5 ⇒ sin ⁡ ( α + β ) = 16 + 9 25 ⇒ sin ⁡ ( α + β ) = 25 25 ⇒ sin ⁡ ( α + β ) = 1 ⇒ sin ⁡ ( α + β ) = sin ⁡ π 2 ⇒ α + β = π 2 Hence proved.

In a △ A B C , A is obtuse. If sin ⁡ A = 3 5 and sin ⁡ B = 5 13 , then show that sin ⁡ C = 16 65 .

Since A is obtuse angle. So, it lies in second quadrant. Now, sin ⁡ A = 3 5 ⇒ cos ⁡ A = 1 − sin 2 ⁡ A ∵ cos 2 x + sin 2 x = 1 = 1 − 3 5 2 = 16 25 = − 4 5 and tan ⁡ A = sin ⁡ A cos ⁡ A ∵ tan x = sin x cos x = 3 5 − 4 5 = − 3 4 Also, sin ⁡ B = 5 13 ⇒ cos ⁡ B = 1 − sin 2 ⁡ B = 1 − 5 13 2 = 144 169 = 12 13 and tan ⁡ B = sin ⁡ B cos ⁡ B = 5 13 12 13 = 5 12 We know that, A + B + C = π ⇒ tan ⁡ A + tan ⁡ B + tan ⁡ C = tan ⁡ A tan ⁡ B tan ⁡ C ⇒ − 3 4 + 5 12 + tan ⁡ C = − 3 4 × 5 12 × tan ⁡ C ⇒ − 9 + 5 12 + tan ⁡ C = − 15 48 tan ⁡ C ⇒ − 4 12 = − 15 48 − 1 tan ⁡ C ⇒ − 1 3 = − 63 48 tan ⁡ C ⇒ tan ⁡ C = 16 63 = Perpendicular ( P ) Base ( B ) ∴Hypotenuse ( H ) = P 2 + B 2 = 16 2 + 63 2 = 256 + 3969 = 4225 = 65 Thus, sin ⁡ C = P H = 16 65 Hence proved.

If A + B + C = π 2 and if none of A , B , C is an odd multiple of π 2 , then prove that cot ⁡ A + cot ⁡ B + cot ⁡ C = cot ⁡ A cot ⁡ B cot ⁡ C .

Given: A + B + C = π 2 ⇒ A + B = π 2 − C ⇒ cot ⁡ ( A + B ) = cot ⁡ π 2 − C ⇒ cot ⁡ A cot ⁡ B − 1 cot ⁡ A + cot ⁡ B = tan ⁡ C ∵ cot π 2 - x = tan x and cot x + y = cot x cot y - 1 cot x + cot y ⇒ cot ⁡ A cot ⁡ B − 1 cot ⁡ A + cot ⁡ B = 1 cot ⁡ C ∵ tan x = 1 cot x ⇒ cot ⁡ C ( cot ⁡ A cot ⁡ B − 1 ) = cot ⁡ A + cot ⁡ B ⇒ cot ⁡ A cot ⁡ B cot ⁡ C − cot ⁡ C = cot ⁡ A + cot ⁡ B ⇒ cot ⁡ A cot ⁡ B cot ⁡ C = cot ⁡ A + cot ⁡ B + cot ⁡ C Hence proved.

E M B I B E

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

Andhra Pradesh Board Solutions for Exercise 3: Exercise 6 (c)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 3: Exercise 6 (c)

Attempt the free practice questions from Exercise 3: Exercise 6 (c) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 3: Exercise 6 (c) with Hints & Solutions

EASY

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q I

Express $\tan θ$ in terms of $\tan α$ , if $\sin (θ + α) = \cos (θ + α)$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q I

If $\tan k ° = \frac{\cos 11^{°} + \sin 11^{°}}{\cos 11^{°} - \sin 11^{°}}$ and $θ$ is in the third quadrant, then find the value of $k$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q I

If $0^{\circ} < A, B < 90^{\circ}$ , such that $\cos A = \frac{5}{13}$ and $\sin B = \frac{4}{5}$ , find the value of $\sin (A - B)$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q II

If $A + B, A$ are acute angles such that $\sin (A + B) = \frac{24}{25} and \tan A = \frac{3}{4}$ , then find the value of $\cos B$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q II

If $\tan α - \tan β = m$ and $\cot α - \cot β = n$ , then prove that $\cot (α - β) = \frac{1}{m} - \frac{1}{n}$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q II

If $\tan (α - β) = \frac{7}{24} and \tan α = \frac{4}{3}$ , where $α$ and $β$ are in the first quadrant prove that $α + β = \frac{π}{2}$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q II

In a $△ A B C, A$ is obtuse. If $\sin A = \frac{3}{5} and \sin B = \frac{5}{13}$ , then show that $\sin C = \frac{16}{65}$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1A>Chapter 6 - Trigonometric Ratios upto Transformations>Exercise 6 (c)>Q III

If $A + B + C = \frac{π}{2}$ and if none of $A, B, C$ is an odd multiple of $\frac{π}{2}$ , then prove that $\cot A + \cot B + \cot C = \cot A \cot B \cot C$ .

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