Q174 / 270 · NDA 2015 (Phase 2) · Mathematics

Consider the following : 1. $\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\frac{\pi}{2}$ 2. $\tan ^{-1} \sqrt{3}+\tan ^{-1} 1=-\tan ^{-1}(2+\sqrt{3})$ which of the above is/are correct?

  • A
    1 only
  • B
    2 only
  • Correct answer
    C
    Both 1 and 2
  • D
    Neither 1 nor 2

Solution

$\begin{aligned} & \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) \\ & \text { If }(-1 \leq x, y \leq 1) \&\left(x^2+y^2 \leq 1\right) \\ & \Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1} \frac{3}{5} \\ & =\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^2}\right] \\ & =\sin ^{-1}\left[\frac{16}{25}+\frac{9}{25}\right] \\ & =\sin ^{-1}(1)=\frac{\pi}{2} \end{aligned}$ $\therefore$ Statement (1) is correct Again- $\begin{aligned} & \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1-x y}\right] \\ & \operatorname{If} ;(x \gt 0),(y \gt 0) \text { and }(x y \gt 1) \\ & \tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)=\pi+\tan ^{-1}\left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right] \\ & =\pi+\tan ^{-1}\left[\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right] \end{aligned}$ $\begin{aligned} & =\pi+\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{-2}\right) \\ & =\pi+\tan ^{-1}[-(2+\sqrt{3})] \\ & =\pi-\tan ^{-1}(2+\sqrt{3}) \quad \because \tan ^{-1}(-x)=-\tan ^{-1} x \end{aligned}$ $\therefore$ Statement (1) is incorrect.

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

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