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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
Question
Prove that $\sin ^ { - 1 } \frac { 8 } { 17 } + \sin ^ { - 1 } \frac { 3 } { 5 } = \tan ^ { - 1 } \frac { 77 } { 36 }.$
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Answer

We need to prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$
Let us consider, $\sin ^{-1} \frac{8}{17}=x$ and $\sin ^{-1} \frac{3}{5}=y, \quad x, y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $\Rightarrow \quad \sin x=\frac{8}{17}$ and $\sin y=\frac{3}{5}$
Now, $\cos ^2 x=1-\sin ^2 x=1-\frac{64}{289}=\frac{225}{289}$
$\Rightarrow \quad \cos x=\sqrt{\frac{225}{289}}=\frac{15}{17}$ [taking positive square root as $\mathrm{x} \quad x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ] and $\cos ^2 y=1-\sin ^2 y=1-\frac{9}{25}=\frac{16}{25}$
$\Rightarrow \quad \cos y=\sqrt{\frac{16}{25}}=\frac{4}{5}$ [taking positive square root as $\left.x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$
Clearly, $\tan \mathrm{x}=\frac{\sin x}{\cos x}=\frac{8}{15}$ and $\tan y=\frac{3}{4}$
$\Rightarrow \quad x=\tan ^{-1} \frac{8}{15}$ and $y=\tan ^{-1} \frac{3}{4}$
Now, in general we can write LHS $=x+y=\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4}$
$=\tan ^{-1}\left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right)$
$=\tan ^{-1}\left(\frac{32+45}{60-24}\right)=\tan ^{-1}\left(\frac{77}{36}\right)=R H S$
Hence Proved.

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