Question

What is ${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)$ equal to?

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Putting

$x=\frac{1}{4}y=\frac{3}{5}$

we get, $xy=\frac{1}{4}\times \frac{3}{5}=\frac{3}{20}(Since,itis<1)$

$ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{\frac{1}{4}+\frac{3}{5}}{1-\frac{1}{4}\times \frac{3}{5}}=\frac{\frac{17}{20}}{\frac{17}{20}}$ $\Rightarrow ta{n}^{-1}1={45}^o=\frac{\pi }{4}$

Therefore, Correct Answer is $B$