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  4. The value of tan-1(1/2)+ tan-1(1/3)+ tan-1(7/8) is

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Q. The value of $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{7}{8}$ is

Inverse Trigonometric Functions

Solution:

Now, $tan^{-1}\left(\frac{1}{2}\right)+tan^{-1}\left(\frac{1}{3}\right)+tan^{-1} \frac{7}{8} $
$= tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}\right)+tan^{-1} \frac{7}{8} $
$ = tan^{-1} \left( \frac{5/6}{ 5/6}\right)+tan^{-1} \frac{7}{8} = tan^{-1}\left(1\right)+tan^{-1} \frac{7}{8}$
$ = tan^{-1} \left(\frac{1+\frac{7}{8}}{1-\frac{7}{8}}\right)$
$= tan^{-1} \left(15\right)$