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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions4 Marks
Question
Prove that $\tan ^{-1}\left(\frac{63}{16}\right)=\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)$
✓

Answer

Suppose, $\sin ^{-1} \frac{5}{13}=\alpha $
$\therefore \sin \alpha=\frac{5}{13}$
$\tan \alpha=\frac{\sin \alpha}{\sqrt{1-\sin ^2 \alpha}}$
$=\frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}$
$=\frac{5}{13} \times \frac{13}{12}=\frac{5}{12}$ and $\beta=\cos ^{-1} \frac{3}{5} $
$\therefore \cos \beta=\frac{3}{5}$
$\sin \beta=\sqrt{1-\cos ^2 \beta}$
$=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\tan \beta=\frac{\sin \beta}{\cos \beta}$
$=\frac{4}{5} \times \frac{5}{3}=\frac{4}{3}$
$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$
$=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}$
$=\frac{15+48}{36-20}=\frac{63}{16}$
$\Rightarrow \alpha+\beta=\tan ^{-1}\left(\frac{63}{16}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)$
$=\tan ^{-1}\left(\frac{63}{16}\right)$
Hence proved.

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