Solve the equation $\cos \left( {{{\tan }^{ – 1}}x} \right) = \sin \left( {{{\cot }^{ – 1}}\frac{3}{4}} \right)$.
[NCERT,Exemplar.2.3,Q.11,Page.36]

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}$


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