Solve the following equations:

    13. 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)

Answers (1)

Given equation 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x);

Using the formula:

\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]

We can write

2\tan^{-1}(\cos x) = \tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ]

\tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \tan^{-1}\left [2cosec x \right ]

So, we can equate;

=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]

=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]

that implies that \cos x = \sin x.

or    \tan x =1      or    x = \frac{\pi}{4}

Hence we have solution x = \frac{\pi}{4}.

 

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