Question

Find $\frac{dy}{dx}ify={\tan }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

Answer 1

R.E.F image

Solution -

$y=ta{n}^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$

$tany=\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}{\sqrt{1+sinx}-\sqrt{1-sinx}(\sqrt{1+sinx}+\sqrt{1-sinx})}$

$tany=\frac{1+cosx}{sinx}$

$secy=\frac{\sqrt{2+2cosx}}{si{n}^{2}x}$

$se{c}^2\frac{dy}{dx}=\frac{-si{n}^{2}x-(1+cosx)cosx}{si{n}^{2}x}$

$\frac{2+2cosx}{si{n}^{2}x}\frac{dy}{dx}=\frac{-si{n}^{2}x-cosx-co{s}^{2}x}{si{n}^{2}x}$

$\frac{dy}{dx}=\frac{-cosx-1}{2+2cosx}=\frac{-1/2(2+2cosx)}{(2+2cosx)}$

$\frac{dy}{dx}=\frac{-1}{2}$