$2\pi-\left(\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{16}{65}\right)$ is equal to:

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Inverse Trigonometric Functions

2 π - (sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{5}{13} + sin^{- 1} \frac{16}{65})

is equal to:

Answer: (A)

Rewrite each inverse sine as an inverse tangent:

sin^{- 1} \frac{4}{5} = tan^{- 1} \frac{4}{3}

sin^{- 1} \frac{5}{13} = tan^{- 1} \frac{5}{12}

and

sin^{- 1} \frac{16}{65} = tan^{- 1} \frac{16}{63}

. Adding the first two,

tan^{- 1} \frac{4}{3} + tan^{- 1} \frac{5}{12} = tan^{- 1} \frac{63}{16}

, and since

\frac{63}{16}

and

\frac{16}{63}

are reciprocals,

tan^{- 1} \frac{63}{16} + tan^{- 1} \frac{16}{63} = \frac{π}{2}

. Therefore the given expression is

2 π - \frac{π}{2} = \frac{3 π}{2}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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