Now L.H.S
= 2tan-1(cos x)
= tan-1\(\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)\)
= tan-1\(\left(\frac{2 \cos x}{\sin ^{2} x}\right)\)
Putting this value in the given equation, we get
tan-1\(\left(\frac{2 \cos x}{\sin ^{2} x}\right)\) = tan-1(2 cosec x)
⇒ \(\frac{2 \cos x}{\sin ^{2} x}\) = 2 cosec x = \(\frac { 2 }{ sin x }\)
⇒ cos x = sin x or tan x = 1
⇒ x = \(\frac { π }{ 4 }\)