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Solve for x : $2\tan^{-1}(\sin x)=\tan^{-1} (2\sec x),0 < x < \large\frac{\pi}{2}$

Toolbox:
• $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$
By taking sinx in the place of x in the above formula of $2tan^{-1}x$ we have
$2tan^{-1}sinx=tan^{-1} \bigg( \frac{2sinx}{1-sin^2x} \bigg)=tan^{-1} (2secx)$
$\Rightarrow\:\frac{2sinx}{cos^2x}=2secx$
$\large\frac{sinx}{cos^2x}=\large\frac{1}{cosx}$
$\Rightarrow\:\large\frac{sinx}{cosx}=1\:\Rightarrow\:\large\:tanx=1$
$\Rightarrow x=\frac{\pi}{4}$
edited Mar 22, 2013