Question

please answer this question​

Answer 1

Question:-

$\sf \large \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} } = - \frac{1}{ \sqrt{cos {}^{2}x } (cosx + sinx)} .$

Given:-

$\sf \large \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} } = - \frac{1}{ \sqrt{cos {}^{2}x } (cosx + sinx)} .$

To Prove:-

$\sf \large \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} } = - \frac{1}{ \sqrt{cos {}^{2}x } (cosx + sinx)} .$

Solution:-

$\sf \large L.H.S. = \sf \large \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} }$

$\sf \large = \frac{1}{2} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} - 1} . \: \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg]$

$\sf \large = \frac{1}{2} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ - \frac{1}{2} } \: \frac{(1 + tanx)( - sec {}^{2}x) - (1 - tanx)sec {}^{2} x }{(1 + tanx) {}^{2} }$

$\sf \large = \frac{1}{2} \frac{ \bigg[1 + tanx \bigg] ^{ \frac{1}{2} } }{ \bigg[1 - tanx \bigg] ^{ \frac{1}{2} } } \: . \: \frac{ \bigg[ - 2 \: sec {}^{2}x \bigg]}{ \bigg[1 + tanx \bigg] ^{2} }$

$\sf \large = - \frac{sec {}^{2}x }{ \sqrt{{[1 + tanx]} [1 - tanx]}[1 + tanx] }$

$\sf \large = - \frac{sec {}^{2}x.cosx }{ \sqrt{{1 - tan {}^{2} x}(sinx + cosx)} }$

$\sf \large = - \frac{secx}{ \sqrt{{ \bigg[ \frac{cos {}^{2} x - sin {}^{2} x}{cos {}^{2} x} } \bigg]} \bigg[cosx + sinx \bigg]} = - \frac{secx.cosx}{ \sqrt{cos2x} [cosx + sinx]}$

$\sf \large = - \frac{1}{ \sqrt{cos2x} (cosx + sinx)} = R.H.S.$

Answer:-

${ \boxed{ \sf \large \color{red}Hence, \sf \large \frac{d}{dx} \bigg[ \frac{1 - tanx}{1 + tanx} \bigg] ^{ \frac{1}{2} } = - \frac{1}{ \sqrt{cos {}^{2}x } (cosx + sinx)} .}}$

${ \boxed { \sf \large \blue{\therefore L.H.S. = R.H.S.}}}$

Hope you have satisfied. ⚘

Answer 2

Step-by-step explanation:

$\begin{aligned} \text { L.H.S. }=& \tt \frac{d}{d x}\left[\frac{1-\tan x}{1+\tan x}\right]^{\frac{1}{2}} \\ \\=& \tt\frac{d}{d x}\left[\frac{\tan \pi / 4-\tan x}{1+\tan x \cdot \tan \pi / 4}\right]^{\frac{1}{2}} \\ \\ =& \tt \frac{d}{d x}\left[\tan \left(\frac{\pi}{4}-x\right)\right]^{1 / 2} \\ \\ =& \tt\frac{1}{2}\left[\tan \left(\frac{\pi}{4}-x\right)\right]^{-1 / 2} \cdot \sec ^{2}\left(\frac{\pi}{4}-x\right) \cdot ( - 1) \\ \\ =& \tt-\frac{\sec ^{2}(\pi / 4-x)}{\sqrt{\tan \left(\frac{\pi}{4}-x\right)}} \\ \\ =& \red{ \boxed{\tt\frac{ - 1}{\sqrt{\cos 2 x} \cdot \cos x} }}\end{aligned}$