Inverse Trigonometric Functions — Class 12 Maths Solution

We have, $\cos \left( {{{\tan }^{ - 1}}x} \right) = \sin \left( {{{\cot }^{ - 1}}\frac{3}{4}} \right)$

$\Rightarrow$ $\cos \left( {{{\cos }^{ - 1}}\frac{1}{{\sqrt {{x^2} + 1} }}} \right) = \sin \left( {{{\sin }^{ - 1}}\frac{4}{5}} \right)$

Let ${\tan ^{ - 1}}x = {\theta _1} \Rightarrow \tan {\theta _1} = \frac{x}{1}$

$\Rightarrow$ $\cos {\theta _1} = \frac{1}{{\sqrt {{x^2} + 1} }} \Rightarrow {\theta _1} = {\cos ^{ - 1}}\frac{1}{{\sqrt {{x^2} + 1} }}$

and ${\cot ^{ - 1}}\frac{3}{4} = {\theta _2} \Rightarrow \cot {\theta _2} = \frac{3}{4}$

$\Rightarrow$ $\sin {\theta _2} = \frac{4}{5} \Rightarrow {\theta _2} = {\sin ^{ - 1}}\frac{4}{5}$

$\Rightarrow$ $\frac{1}{{\sqrt {{x^2} + 1} }} = \frac{4}{5}$
. and $\left. {\sin \left( {{{\sin }^{ - 1}}x} \right) = x,x \in [ - 1,1]} \right\}$

On squaring both sides,

we get
$16\left( {{x^2} + 1} \right) = 25$
$\Rightarrow$ $16{x^2} = 9$
$\Rightarrow$ ${x^2} = {\left( {\frac{3}{4}} \right)^2}$

$therefore,$ $x = \pm \frac{3}{4} = \frac{{ - 3}}{4},\frac{3}{4}$

LONG ANSWER (L.A.)