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

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