Question

Let tan-1 x ∈($-\frac π2$ $\frac π2$) for x ∈ R. Then the number of real solutions of the equation $\sqrt{1 + \cos (2x)} = \sqrt2 \tan^{-1}(\tan x)$ in the set $(-\frac{3π}{2}, -\frac{π}{2}) ∪ (-\frac{π}{2},\frac{π}{2}) ∪ (\frac{π}{2},\frac{3π}{2})$ is equal to

Answer 1

Given :
$\sqrt{1 + \cos (2x)} = \sqrt2 \tan^{-1}(\tan x)$
$⇒|\cos x|=\tan^{-1}(\tan x)$

Number of solutions = Number of intersection points = 3.
So, the correct answer is 3.