Question

The value of $cot^{-1}\,9 + cosec^{-1} \frac{\sqrt{41}}{4}$ is given by

• A. $0$
• B. $\frac{\pi}{4}$
• C. $tan^{-1}\,2$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (B)

$cot^{-1}\,9 + cosec^{-1} \frac{\sqrt{41}}{4}$
$= cot^{-1}\,9 + cosec^{-1} \frac{4}{\sqrt{41}}$
$cot^{-1}\,9 + cot^{-1} \frac{5}{4} $
$= tan^{-1} \frac{1}{9} + tan^{-1} \frac{4}{5}$
$= tan^{-1}\left(\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9} \times\frac{4}{5}}\right)$
$= tan^{-1} \left(1\right) = \frac{\pi}{4}$