Question

$2{\cot }^{-1}7+{\cos }^{-1}\frac{3}{5}$ is equal to

• A. ${\cot }^{-1}\left(\frac{44}{117}\right)$
• B. $cose{c}^{-1}\left(\frac{125}{117}\right)$
• C. ${\tan }^{-1}\left(\frac{4}{117}\right)$
• D. ${\tan }^{-1}\left(\frac{44}{125}\right)$

Answer 1

The correct option is A ${\cot }^{-1}\left(\frac{44}{117}\right)$

$2{\cot }^{-1}7+{\cos }^{-1}\left(\frac{3}{5}\right)$

Let ${\cos }^{-1}\left(\frac{3}{5}\right)=\theta$

$\cos \theta =\frac{3}{5}$

$\cot \theta =\frac{3}{4}$

$2{\cot }^{-1}\left(7\right)+{\cot }^{-1}\left(\frac{3}{4}\right)$ $[\because 2{\cot }^{-1}(x)={\cot }^{-1}\left(\frac{x^2-1}{2x}\right)]$

${\cot }^{-1}\left(\frac{7^2-1}{2\times 7}\right)+{\cot }^{-1}\left(\frac{3}{4}\right)$

${\cot }^{-1}\left(\frac{24}{7}\right)+{\cot }^{-1}\left(\frac{3}{4}\right)$

${\cot }^{-1}\left(\frac{\frac{24}{7}\times \frac{3}{4}-1}{\frac{24}{7}+\frac{3}{4}}\right)$ $[\because {\cot }^{-1}(a)+{\cot }^{-1}(b)={\cot }^{-1}\left(\frac{ab-1}{a+b}\right)]$

${\cot }^{-1}\left(\frac{44/28}{117/28}\right)={\cot }^{-1}\left(\frac{44}{117}\right)$.