सरल कीजिए: $2 \tan^{-1} (\cos x) = \tan^{-1} (2 \operatorname{cosec} x)$
Miscellaneous Exercise-13
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ज्ञात है, $2 \tan^{-1} (\cos x) = \tan^{-1} (2 \operatorname{cosec} x)$
$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right) = \tan ^{-1}(2 \operatorname{cosec} x)[\because 2 \tan^{-1} x = \tan^{-1} \left(\frac{2 x}{1-x^{2}}\right)]$
$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right) = \tan ^{-1}\left(\frac{2}{\sin x}\right) \Rightarrow \frac{2 \cos x}{\sin ^{2} x} = \frac{2}{\sin x} \Rightarrow \frac{2 \cos x}{\sin x} = 2$
$\Rightarrow \cot x = 1 \Rightarrow \cot x = \cot \frac{\pi}{4} \Rightarrow x = \frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right) = \tan ^{-1}(2 \operatorname{cosec} x)[\because 2 \tan^{-1} x = \tan^{-1} \left(\frac{2 x}{1-x^{2}}\right)]$
$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right) = \tan ^{-1}\left(\frac{2}{\sin x}\right) \Rightarrow \frac{2 \cos x}{\sin ^{2} x} = \frac{2}{\sin x} \Rightarrow \frac{2 \cos x}{\sin x} = 2$
$\Rightarrow \cot x = 1 \Rightarrow \cot x = \cot \frac{\pi}{4} \Rightarrow x = \frac{\pi}{4}$
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