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MHT CET · Maths · Inverse Trigonometric Functions

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

  1. A \(\frac{\pi}{2}\)
  2. B \(\frac{5 \pi}{4}\)
  3. C \(\frac{7 \pi}{4}\)
  4. D \(\frac{3 \pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{3 \pi}{2}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & 2 \pi-\left[\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =2 \pi-\left[\left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =2 \pi-\left[\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =2 \pi-\left[\tan ^{-1}\left(\frac{63}{16}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =2 \pi-\left[\cos ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =2 \pi-\frac{\pi}{2} \\ & =\frac{3 \pi}{2}\end{aligned}\)