Question

The value of $co{t}^{-1}9+cose{c}^{-1}\frac{\sqrt{41}}{4}$ is given by

• A. 0
• B. π/4
• C. tan^-12
• D. π/2

Answer 1

The correct option is B π/4

$co{t}^{-1}9+cose{c}^{-1}\frac{\sqrt{41}}{4}$
$=co{t}^{-1}9+si{n}^{-1}\frac{\sqrt{41}}{4}$
$=co{t}^{-1}9ta{n}^{-1}\frac{4}{5}$
$=ta{n}^{-1}\left(\frac{1}{9}\right)+ta{n}^{-1}\left(\frac{4}{5}\right)$
$=ta{n}^{-1}\left|\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9}\times \frac{4}{5}}\right|$
$=ta{n}^{-1}\left(\frac{5+36}{45-4}\right)=ta{n}^{-1}\left(\frac{41}{41}\right)=ta{n}^{-1}\left(1\right)$
$=\frac{\pi }{4}$