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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions1 Mark
Question
Prove that: ${\tan ^{ - 1}}\frac{{63}}{{16}} = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$
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Answer

Let ${\sin ^{ - 1}}\frac{5}{{13}} = \theta$ so that $\sin \theta = \frac{5}{{13}}$
$\therefore \cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \frac{{25}}{{169}}}$$= \sqrt {\frac{{144}}{{169}}} = \frac{{12}}{{13}}$
$\therefore \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{5}{{12}}$
Again, Let ${\cos ^{ - 1}}\frac{3}{5} = \phi$ so that $\cos \phi = \frac{3}{5}$
$\therefore \sin \phi = \sqrt {1 - {{\cos }^2}\phi } = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$\therefore \tan \phi = \frac{{\sin \phi }}{{\cos \phi }} = \frac{4}{3}$
Since $\tan \left( {\theta + \phi } \right) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \times \frac{4}{3}}}$
$= \frac{{\frac{{21}}{{12}}}}{{\frac{4}{9}}} = \frac{{63}}{{16}}$
$\Rightarrow \theta + \phi = {\tan ^{ - 1}}\frac{{63}}{{16}}$
$\Rightarrow {\tan ^{ - 1}}\frac{{63}}{{16}} = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$

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