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  4. If angle A = 90° in the Δ ABC, then tan-1((c/a + b)) + tan-1 ((b/a + c)) is equal to

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Q. If $\angle A = 90^°$ in the $\Delta ABC$, then $\tan^{-1}\left(\frac{c}{a + b}\right) + \tan^{-1} \left(\frac{b}{a + c}\right)$ is equal to

Inverse Trigonometric Functions

Solution:

We have, $ABC$ is a right angled $\Delta$ at $A$
$\Rightarrow \,\,a^{2} = b^{2} + c^{2}\ldots\left(i\right)$
Now, $\tan^{-1}\left(\frac{c}{a + b}\right) + \tan^{-1} \left(\frac{b}{a + c}\right)$
$ = \tan^{-1}\left\{\frac{\left(\frac{c}{a + b}\right) +\left(\frac{b}{a + c}\right)}{1-\left(\frac{c}{a + b}\right) \left(\frac{b}{a + c}\right)}\right\}$
image
$ = \tan^{-1}\left\{\frac{ac+c^{2} +b^{2} +ab}{a^{2} + ab+ac}\right\}$
$ = \tan^{-1}\left\{\frac{ac+a^{2} +ab}{a^{2} + ab+ac}\right\}$ using $\left(i\right)$
$= \tan^{-1}\left(1\right) = \frac{\pi}{4}$