Question

In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]

 

 

Options

• `pi/3`

• `pi/4`

• `(5x)/2`

• `pi/6`

Answer 1

(b) `pi/4`

We know
\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[\therefore \tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) = \tan^{- 1} \left( \frac{\frac{a}{b + c} + \frac{b}{c + a}}{1 - \frac{a}{b + c} \times \frac{b}{c + a}} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{ac + a^2 + b^2 + bc}{\left( b + c \right)\left( c + a \right)}}{\frac{ac + c^2 + bc}{\left( b + c \right)\left( c + a \right)}} \right)\]
\[= \tan^{- 1} \left( \frac{ac + c^2 + bc}{ac + c^2 + bc} \right) \left[ \because a^2 + b^2 = c^2 \right]\]
\[ = \tan^{- 1} \left( 1 \right)\]
\[ = \tan^{- 1} \left( \tan\frac{\pi}{4} \right)\]
\[ = \frac{\pi}{4}\]