2 tan^{-1}(cos x) = tan^{-1}(csc^2 x) હોય,તો | vedclass

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(D) આપેલ સમીકરણ: $2 	an^{-1}(\cos x) = 	an^{-1}(\csc^2 x)$.સૂત્ર $2 	an^{-1}(	heta) = 	an^{-1}\left(\frac{2	heta}{1-	heta^2}ight)$ નો ઉપયોગ ક

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$\frac{\pi}{3}$

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(D) આપેલ સમીકરણ: $2 	an^{-1}(\cos x) = 	an^{-1}(\csc^2 x)$.સૂત્ર $2 	an^{-1}(	heta) = 	an^{-1}\left(\frac{2	heta}{1-	heta^2}ight)$ નો ઉપયોગ કરતા:$	an^{-1}\left(\frac{2 \cos x}{1 - \cos^2 x}ight) = 	an^{-1}(\csc^2 x)$.આપણે જાણીએ છીએ કે $1 - \cos^2 x = \sin^2 x$,તેથી:$\frac{2 \cos x}{\sin^2 x} = \frac{1}{\sin^2 x}$.જો $\sin^2 x 
eq 0$ હોય,તો બંને બાજુથી $\sin^2 x$ ને દૂર કરતા:$2 \cos x = 1$.$\cos x = \frac{1}{2}$.આમ,$x = \frac{\pi}{3}$.

$2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$ હોય,તો $x = $

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