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 $\frac{dy}{dx}=\frac{1}{a+b\cos x^{\prime }}a>b>0$

Answer 1

$\frac{dy}{dx}=\frac{2\sqrt{\frac{a-b}{a+b}}\ast \left(\frac{se{c}^2\frac{x}{2}}{2}\right)}{\sqrt{a^2-b^2}\ast (1+\frac{(a-b)ta{n}^2\frac{x}{2}}{a+b})}$

$\frac{dy}{dx}=\frac{se{c}^2\frac{x}{2}}{(a+b)\ast (1+\frac{(a-b)ta{n}^2\frac{x}{2}}{a+b})}$

$\frac{dy}{dx}=\frac{se{c}^2\frac{x}{2}}{(a+b)+(a-b)ta{n}^2\frac{x}{2}}$

$\frac{dy}{dx}=\frac{se{c}^2\frac{x}{2}}{a(1+ta{n}^2\frac{x}{2})+b(1-ta{n}^2\frac{x}{2})}$

$\frac{dy}{dx}=\frac{1}{a(co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2})+b(co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2})}$

$\frac{dy}{dx}=\frac{1}{a+bcosx}$