Question

If the value of $cot\left({\left(cosec\right)}^{-1}\frac{5}{3}+{\left(tan\right)}^{-1}\frac{2}{3}\right)$ is $\frac{k}{17}$ then find $k$ .

Answer 1

Answer: 6

We have, $cot\left[cose{c}^{-1}\frac{5}{3}+ta{n}^{-1}\frac{2}{3}\right]$
$\because cose{c}^{-1}\frac{5}{3}=ta{n}^{-1}\frac{3}{4}$ (with the help of right-angle triangle)
Then, $cot\left[ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{2}{3}\right]$
$=cot\left[{\left(tan\right)}^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}\right)\right]$
$=cot\left[{\left(tan\right)}^{-1}\left(\frac{\frac{9+8}{12}}{\frac{12-6}{12}}\right)\right]$
$=cot\left[ta{n}^{-1}\frac{17}{6}\right]$
$=cot\left[{\left(cot\right)}^{-1}\frac{6}{17}\right],\left(\because {\left(cot\right)}^{-1}x={\left(tan\right)}^{-1}\frac{1}{x},x>0\right)$
$=\frac{6}{17}$ $\left(\because cot\left({\left(cot\right)}^{-1}x\right)=x\right)$
$\therefore k=6$