The value of $\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{3}{5}\right)$ is equal to :

(D) $\frac{\pi}{2}-\sin ^{-1}\left(\frac{56}{65}\right)$

(A) $\pi-\cos ^{-1}\left(\frac{33}{65}\right)$

(B) $\pi-\sin ^{-1}\left(\frac{63}{65}\right)$

(C) $\frac{\pi}{2}-\cos ^{-1}\left(\frac{9}{65}\right)$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Inverse Trigonometric Functions

The value of

sin^{- 1} (\frac{12}{13}) - sin^{- 1} (\frac{3}{5})

is equal to :

Answer: (D)

Use

sin^{- 1} x - sin^{- 1} y = sin^{- 1} (x 1 - y^{2} - y 1 - x^{2})

with

x = \frac{12}{13}

and

y = \frac{3}{5}

. Here

1 - (12/13)^{2} = \frac{5}{13}

and

1 - (3/5)^{2} = \frac{4}{5}

, so the expression is

sin^{- 1} (\frac{12}{13} \cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{5}{13}) = sin^{- 1} (\frac{48 - 15}{65}) = sin^{- 1} \frac{33}{65}

. Writing this as

\frac{π}{2} - cos^{- 1} \frac{33}{65}

and noting

cos^{- 1} \frac{33}{65} = sin^{- 1} \frac{56}{65}

(since

3 3^{2} + 5 6^{2} = 6 5^{2}

), the value is

\frac{π}{2} - sin^{- 1} \frac{56}{65}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option