Question

Solve the following equation `cos(tan^-1x) = sin(cot^-1  3/4)`

Answer 1

Given that `cos(tan^-1x) = sin(cot^-1  3/4)`

⇒ `cos[cos^-1  1/sqrt(1 + x^2)] = sin[sin^-1  4/5]`  ......

\begin{bmatrix}\begin{array}[b]{r} \left( \because \tan^{-1}x = \cos^{-1} (\frac{1}{\sqrt{1 + x^2}} \right)\\ \left( \cot^{-1}x = \sin^{-1} (\frac{1}{\sqrt{1 + x^2}}\right) \end{array}\end{bmatrix}

⇒  `1/sqrt(1 + x^2) = 4/5`

Squaring both sides we get,

`1/(1 + x^2) = 16/25`

⇒ `1 + x^2 = 25/16`

⇒ `x^2 = 25/16 - 1 = 9/16`

⇒  x = `+- 3/4`

Hence, x = `(-3)/4, 3/4`.