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Question

If $$y = \frac{2}{{\sqrt {{a^2} - {b^2}} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right)$$, prove that $$\frac{{dy}}{{dx}} = \frac{1}{{a + b\cos x'}}a > b > 0$$


Solution

$$\cfrac{dy}{dx} = \cfrac{2 \sqrt{\cfrac{a-b}{a+b}} *(\cfrac{sec^2 \cfrac{x}{2}}{2})}{\sqrt{a^2-b^2}*(1+\cfrac{(a-b)tan^2 \cfrac{x}{2}}{a+b})}$$
$$\cfrac{dy}{dx} = \cfrac{sec^2 \cfrac{x}{2}}{{(a+b)}*(1+\cfrac{(a-b)tan^2 \cfrac{x}{2}}{a+b})}$$
$$\cfrac{dy}{dx} = \cfrac{sec^2 \cfrac{x}{2}}{(a+b)+(a-b)tan^2 \cfrac{x}{2}}$$
$$\cfrac{dy}{dx} = \cfrac{sec^2 \cfrac{x}{2}}{a(1+tan^2 \cfrac{x}{2})+b(1-tan^2\cfrac{x}{2})}$$
$$\cfrac{dy}{dx} = \cfrac{1}{a(cos^2 \cfrac{x}{2}+sin^2 \cfrac{x}{2})+b(cos^2 \cfrac{x}{2}-sin^2\cfrac{x}{2})}$$
$$\cfrac{dy}{dx} = \cfrac{1}{a+bcosx}$$

Mathematics

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