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Consider the following :
1) $\quad \sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\frac{\pi}{2}$
2) $\tan ^{-1} \sqrt{3}+\tan ^{-1} 1=-\tan ^{-1}(2+\sqrt{3})$
which of the above is/are correct?
Options:
1) $\quad \sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\frac{\pi}{2}$
2) $\tan ^{-1} \sqrt{3}+\tan ^{-1} 1=-\tan ^{-1}(2+\sqrt{3})$
which of the above is/are correct?
Solution:
2841 Upvotes
Verified Answer
The correct answer is:
1 only
$\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)$
$\operatorname{If}(-1 \leq x, y \leq 1) \&\left(x^{2}+y^{2} \leq 1\right)$
$\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1} \frac{3}{5}$
$=\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right]$
$=\sin ^{-1}\left[\frac{16}{25}+\frac{9}{25}\right]=\sin ^{-1}(1)=\frac{\pi}{2}$
$\therefore$ Statement $(1)$ is correct
Again, $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1-x y}\right]$
$\begin{aligned} \text { If } ;(x>0),(y>0) \text { and }(x y>1) \\ \tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)=\pi+\tan ^{-1}\left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right] \\=\pi+\tan ^{-1}\left[\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right] \\=\pi+\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{-2}\right)=\pi+\tan ^{-1}[-(2+\sqrt{3})] \\=& \pi-\tan ^{-1}(2+\sqrt{3}) \quad \because \tan ^{-1}(-x)=-\tan ^{-1} x \end{aligned}$
$\therefore$ Statement $(2)$ is incorrect
$\operatorname{If}(-1 \leq x, y \leq 1) \&\left(x^{2}+y^{2} \leq 1\right)$
$\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1} \frac{3}{5}$
$=\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right]$
$=\sin ^{-1}\left[\frac{16}{25}+\frac{9}{25}\right]=\sin ^{-1}(1)=\frac{\pi}{2}$
$\therefore$ Statement $(1)$ is correct
Again, $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1-x y}\right]$
$\begin{aligned} \text { If } ;(x>0),(y>0) \text { and }(x y>1) \\ \tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)=\pi+\tan ^{-1}\left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right] \\=\pi+\tan ^{-1}\left[\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right] \\=\pi+\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{-2}\right)=\pi+\tan ^{-1}[-(2+\sqrt{3})] \\=& \pi-\tan ^{-1}(2+\sqrt{3}) \quad \because \tan ^{-1}(-x)=-\tan ^{-1} x \end{aligned}$
$\therefore$ Statement $(2)$ is incorrect
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