Differentiate ${tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $tan^-1((sinx-cosx)/(sinx+cosx))$ with respect to $\frac{x}{2}$ $x/2$ ?

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Differentiate ${tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $tan^-1((sinx-cosx)/(sinx+cosx))$ with respect to $\frac{x}{2}$ $x/2$ ?

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Answer 1 · 2018-04-07T07:34:15

Let $y = {tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $y=tan^-1((sinx-cosx)/(sinx+cosx))$

$\Rightarrow {tan}^{- 1} (\frac{\frac{sin x}{cos x} - \frac{cos x}{cos x}}{\frac{sin x}{cos x} + \frac{cos x}{cos x}})$ $=>tan^-1((sinx/cosx-cosx/cosx)/(sinx/cosx+cosx/cosx))$

$\Rightarrow {tan}^{- 1} (\frac{tan x - 1}{tan x + 1})$ $=>tan^-1((tanx-1)/(tanx+1))$

$\Rightarrow {tan}^{- 1} (\frac{tan x - {tan 45}^{\circ}}{1 + tan x {tan 45}^{\circ}})$ $=>tan^-1((tanx-tan45^@)/(1+tanxtan45^@))$

$\Rightarrow {tan}^{- 1} ((tan (x - 45^{\circ}))$ $=>tan^-1((tan(x-45^@))$

$\Rightarrow x - \frac{π}{4}$ $=> x-pi/4$

Therefore, $\frac{d y}{d x} = 1$ $dy/dx= 1$

Now. let $z = \frac{x}{2}$ $z=x/2$

Therefore, $\frac{d z}{d x} = \frac{1}{2}$ $(dz)/dx = 1/2$

The question says, "Differentiate ${tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $tan^-1((sinx-cosx)/(sinx+cosx))$ with respect to $\frac{x}{2}$ $x/2$ "

$\Rightarrow \frac{d y}{d z} = \frac{d y}{d x} \times \frac{d x}{d z}$ $=> dy/dz = dy/dx xx dx/dz$

$\Rightarrow 1 \times \frac{2}{1} = 2$ $=> 1 xx 2/1 =2$

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Differentiate ${tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $tan^-1((sinx-cosx)/(sinx+cosx))$ with respect to $\frac{x}{2}$ $x/2$ ?

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Differentiate ${tan}^{- 1} (\frac{sin x - cos x}{sin x + cos x})$ $tan^-1((sinx-cosx)/(sinx+cosx))$ with respect to $\frac{x}{2}$ $x/2$ ?