Question

The value of ${\cot }^{-1}9+\cos e{c}^{-1}\frac{\sqrt{41}}{4}$

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

Answer 1

The Correct Option is B

Explanation:
${\cot }^{-1}9+{\mathrm{cosec}}^{-1}\frac{\sqrt{41}}{4}$$={\cot }^{-1}9+{\sin }^{-1}\frac{4}{\sqrt{41}}$$={\cot }^{-1}9+{\tan }^{-1}\frac{4}{5}$$={\tan }^{-1}\left(\frac{1}{9}\right)+{\tan }^{-1}\left(\frac{4}{5}\right)$$={\tan }^{-1}\left|\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9}\times \frac{4}{5}}\right|$$={\tan }^{-1}\left(\frac{5+36}{45-4}\right)={\tan }^{-1}\left(\frac{41}{41}\right)={\tan }^{-1}(1)$$=\frac{\pi }{4}$