Question

If ${\tan }^{-1}\frac{b}{c+a}+{\tan }^{-1}\frac{c}{a+b}=\frac{\pi }{4},$ where $a,b,c$ are the sides of $△ABC,$ then $△ABC$ is

• A. acute angled triangle
• B. obtuse angled triangle
• C. right angled triangle
• D. equilateral triangle

Answer 1

The correct option is C right angled triangle

Given, ${\tan }^{-1}\frac{b}{c+a}+{\tan }^{-1}\frac{c}{a+b}=\frac{\pi }{4}$
We know that the sum of any two sides of a triangle is greater than the third side.
$\therefore \frac{b}{c+a}<1$ and $\frac{c}{a+b}<1$
$\Rightarrow \frac{b}{c+a}\cdot \frac{c}{a+b}<1$

Hence, the given equation reduces to
${\tan }^{-1}\left(\frac{\frac{b}{c+a}+\frac{c}{a+b}}{1-\frac{b}{c+a}\cdot \frac{c}{a+b}}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{ab+b^2+c^2+ac}{ac+bc+a^2+ab-bc}=1$
$\Rightarrow ab+b^2+c^2+ac=ac+a^2+ab$
$\Rightarrow b^2+c^2=a^2$
$\therefore △ABC$ is right angled at $A.$