Search any question & find its solution
Question:
Answered & Verified by Expert
If in a $\triangle A B C, \angle C=\frac{\pi}{2}$, then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$
Options:
Solution:
2792 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Given, $\angle C=\frac{\pi}{2} \Rightarrow c^{2}=a^{2}+b^{2}$
\{by pythagoras $\}$
Now, $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$
$=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]$
$=\tan ^{-1}\left[\frac{a c+a^{2}+b^{2}+b c}{b c+b a+c^{2}+c a-a b}\right)$
$=\tan ^{-1}\left(\frac{a c+c^{2}+b c}{b c+c^{2}+c a}\right)$
$=\tan ^{-1}(1)=\frac{\pi}{4}$
\{by pythagoras $\}$
Now, $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$
$=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]$
$=\tan ^{-1}\left[\frac{a c+a^{2}+b^{2}+b c}{b c+b a+c^{2}+c a-a b}\right)$
$=\tan ^{-1}\left(\frac{a c+c^{2}+b c}{b c+c^{2}+c a}\right)$
$=\tan ^{-1}(1)=\frac{\pi}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.