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  4. 7 In triangle ABC if angle A =(π/2), then tan -1 ( b / c + a )+ tan -1 ( c / a + b ) equals

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Q. 7 In $\triangle ABC$ if $\angle A =\frac{\pi}{2}$, then $\tan ^{-1} \frac{ b }{ c + a }+\tan ^{-1} \frac{ c }{ a + b }$ equals

Inverse Trigonometric Functions

Solution:

$\tan ^{-1} \frac{b}{c+a}+\tan ^{-1} \frac{c}{a+b}=\tan ^{-1}\left(\frac{\frac{b}{c+a}+\frac{c}{a+b}}{1-\frac{b c}{(c+a)(a+b)}}\right)$
$=\tan ^{-1}\left(\frac{a b+c a+b^2+c^2}{c a+a b+a^2}\right)=\tan ^{-1} 1=\frac{\pi}{4}$