If (cosec 2x + cot 2x) (5-2\(\sqrt {6}\))=1, find the value of cosx

\(\sqrt {\frac{\sqrt {6}}{5} + \frac{1}{2}}\)

We know cosec²θ - cot²θ = 1

(cosecθ + cotθ)(cosec θ - cotθ) = 1

⇒ cosec 2x + cot 2x = \(\frac{1}{5-2 \sqrt {6}}\)

⇒ cosec 2x - cot 2x = 5-2\(\sqrt {6}\)

2 cosec 2x = 10

cosec 2x = \(\frac {5}{1}\)

we know ⇒ cos 2x = \(\frac{2\sqrt {6}}{5}\)

1 - 2sin²x = 2cos²x - 1 = \(\frac{2\sqrt {6}}{5}\)

2cos²x - 1 = \(\frac{2\sqrt {6}}{5}\)

2cos²x - 1 = \(\frac{2\sqrt {6}+5}{5}\)

cosx = \(\sqrt {\frac{2\sqrt {6}+5}{5}}\) = \(\sqrt {\frac{\sqrt {6}}{5} + \frac{1}{2}}\)

= \(\sqrt {\frac{\sqrt {6}}{5} + \frac{1}{2}}\)