The number of real solutions $x$ of the equation $\cos ^ { 2 } ( x \sin ( 2 x ) ) + \frac { 1 } { 1 + x ^ { 2 } } - \cos ^ { 2 } x + \sec ^ { 2 } x$ is-

$\begin{array}{l} \text { L.H.S }= \cos ^{2}(x \sin (2 x))+\frac{1}{1+x^{2}} \\ -1 \leq \cos x \leq 1 \\ 0 \leq \cos ^{2} x \leq 1 \\ 0 \leq \cos ^{2}(x \sin 2 x) \leq 1 \end{array}$

$1 \geq \frac{1}{1+x^{2}}>0 \quad[\because$ it is a proper fraction $]$

$\text { R.H.S }=\cos ^{2} x+\sec ^{2} x$

Maximum value of $\cos ^{2} x=1$

Minimum value of $\sec ^{2} x=1$

$\therefore \cos ^{2} x+\sec ^{2} x \geqslant 2$

$\therefore$ Only solution will be at $x=0$

$\begin{array}{l} \text { L.H.S }=1+1=2 \\ \text { R.H.S }=1+1=2 \end{array}$

$\therefore$ Number of solutions $=1$