Question

The value of $\cot \left(cose{c}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\right)$ is

• A. $\frac{5}{17}$
• B. $\frac{6}{17}$
• C. $\frac{3}{17}$
• D. $\frac{4}{17}$

Answer 1

Answer: (B)

$cose{c}^{-1}\frac{5}{3}=si{n}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{3}{4}$
$\therefore cot\left(cose{c}^{-1}\frac{5}{3}+ta{n}^{-1}\frac{2}{3}\right)$
$=cot\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{2}{3}\right)$
$=cot\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)$
$=cot\left(ta{n}^{-1}\frac{\frac{17}{12}}{\frac{1}{2}}\right)$
$=cot(ta{n}^{-1}\frac{17}{6})$
$=cot\left(co{t}^{-1}\frac{6}{17}\right)$
$=\frac{6}{17}$