Q215 / 270 · NDA 2014 (Phase 2) · Mathematics

What is $\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}$ equal to?

  • Correct answer
    A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Solution

Let $\sin ^{-1} \frac{3}{5}=x$ $\Rightarrow \quad \sin x=\frac{3}{5} \qquad \therefore \cos x=\frac{4}{5}$ and $\quad \sin ^{-1} \frac{4}{5}=y$ $\Rightarrow \quad \sin y=\frac{4}{5} \qquad \therefore \cos y=\frac{3}{5}$ $\begin{aligned} \sin (x+y) & =\sin x \cdot \cos y+\cos x \cdot \sin y \\ & =\frac{3}{5} \times \frac{3}{5}+\frac{4}{5} \times \frac{4}{5}=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1 \\ \therefore \quad \sin (x+y) & =\sin \frac{\pi}{2} \\ \Rightarrow \quad x+y & =\frac{\pi}{2} \\ \Rightarrow & \sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2} .\end{aligned}$

Hence, the correct answer is option: 'A'.

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