If $y = \large\frac{2}{\sqrt{a^2-b^2}}$$\tan^{-1} \bigg( \sqrt{\frac{a-b}{a+b}} \tan\large\frac{x}{2} \bigg).$ prove that $\large\frac{dy}{dx} =\large \frac{1}{(a+bcos \: x)}.$

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