सिद्ध कीजिए: $\tan ^{-1} \frac{63}{16}$ = $\sin ^{-1} \frac{5}{13}$ + $\cos ^{-1} \frac{3}{5}$
Miscellaneous Exercise-7
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ज्ञात है,$ \tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1 }\frac{3}{5}$
दायाँ पक्ष $= \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
मान लीजिए $\sin^{-1} \left(\frac{5}{13}\right) $
$= x \Rightarrow \sin x =\frac{5}{13}$
$\because 1-\sin ^{2} x = \cos ^{2} x $
$ \Rightarrow 1-\left(\frac{5}{13}\right)^{2} = \cos ^{2} x$
$\Rightarrow \frac{169-25}{169} = \cos ^{2} x $
$ \Rightarrow \frac{144}{169} = \cos ^{2} x $
$ \Rightarrow \frac{12}{13} = \cos x$
$\therefore \tan x = \frac{\sin x}{\cos x}$
$ \Rightarrow \tan x = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
$ \Rightarrow x = \tan^{-1}\frac{5}{12}$
माना $\cos ^{-1} \frac{3}{5} = y $
$\Rightarrow \cos y = \frac{3}{5}$
$\because 1-\cos ^{2} y = \sin ^{2} y$
$ \Rightarrow 1 - \left(\frac{3}{5}\right)^{2} = \sin ^{2} y$
$\Rightarrow \frac{25-9}{25}= \sin ^{2} y$
$ \Rightarrow \frac{16}{25} = \sin ^{2} y$
$ \Rightarrow \frac{4}{5} = \sin y$
$\therefore \tan y = \frac{\sin y}{\cos y}$
$ \Rightarrow \tan y = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$
$ \Rightarrow y = \tan ^{-1} \frac{4}{3}$
अत: ज्ञात समीकरण का प्रारूप इस प्रकार है
$\tan ^{-1} \frac{63}{16} = x + y$
$ \Rightarrow \tan ^{-1} \frac{63}{16} = \tan ^{-1}\left(\frac{5}{12}\right) + \tan ^{-1}\left(\frac{4}{3}\right)$
$\therefore$ दायाँ पक्ष $= \tan ^{-1}\left(\frac{5}{12}\right) + \tan ^{-1}\left(\frac{4}{3}\right) = \tan ^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right) [ \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)]$
$= \tan ^{-1}\left(\frac{\frac{15+48}{12 \times 3}}{\frac{12 \times 3-20}{12 \times 3}}\right) = \tan ^{-1}\left(\frac{63}{36} \times \frac{36}{36-20}\right) = \tan ^{-1}\left(\frac{63}{16}\right) =$ बायाँ पक्ष
$\therefore$ बायाँ पक्ष $=$ दायाँ पक्ष इति सिद्धम्
दायाँ पक्ष $= \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
मान लीजिए $\sin^{-1} \left(\frac{5}{13}\right) $
$= x \Rightarrow \sin x =\frac{5}{13}$
$\because 1-\sin ^{2} x = \cos ^{2} x $
$ \Rightarrow 1-\left(\frac{5}{13}\right)^{2} = \cos ^{2} x$
$\Rightarrow \frac{169-25}{169} = \cos ^{2} x $
$ \Rightarrow \frac{144}{169} = \cos ^{2} x $
$ \Rightarrow \frac{12}{13} = \cos x$
$\therefore \tan x = \frac{\sin x}{\cos x}$
$ \Rightarrow \tan x = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
$ \Rightarrow x = \tan^{-1}\frac{5}{12}$
माना $\cos ^{-1} \frac{3}{5} = y $
$\Rightarrow \cos y = \frac{3}{5}$
$\because 1-\cos ^{2} y = \sin ^{2} y$
$ \Rightarrow 1 - \left(\frac{3}{5}\right)^{2} = \sin ^{2} y$
$\Rightarrow \frac{25-9}{25}= \sin ^{2} y$
$ \Rightarrow \frac{16}{25} = \sin ^{2} y$
$ \Rightarrow \frac{4}{5} = \sin y$
$\therefore \tan y = \frac{\sin y}{\cos y}$
$ \Rightarrow \tan y = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$
$ \Rightarrow y = \tan ^{-1} \frac{4}{3}$
अत: ज्ञात समीकरण का प्रारूप इस प्रकार है
$\tan ^{-1} \frac{63}{16} = x + y$
$ \Rightarrow \tan ^{-1} \frac{63}{16} = \tan ^{-1}\left(\frac{5}{12}\right) + \tan ^{-1}\left(\frac{4}{3}\right)$
$\therefore$ दायाँ पक्ष $= \tan ^{-1}\left(\frac{5}{12}\right) + \tan ^{-1}\left(\frac{4}{3}\right) = \tan ^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right) [ \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)]$
$= \tan ^{-1}\left(\frac{\frac{15+48}{12 \times 3}}{\frac{12 \times 3-20}{12 \times 3}}\right) = \tan ^{-1}\left(\frac{63}{36} \times \frac{36}{36-20}\right) = \tan ^{-1}\left(\frac{63}{16}\right) =$ बायाँ पक्ष
$\therefore$ बायाँ पक्ष $=$ दायाँ पक्ष इति सिद्धम्
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