Question

differentiate
{ tan^-1 cosx/1+sinx}​

Answer 1

Given,

$\longrightarrow y=\tan^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)$

We've to differentiate this $y$ wrt $x.$

We take the tangent and,

$\longrightarrow \dfrac{\cos x}{1+\sin x}=\tan y$

But,

• $\cos(2A)=\cos^2A-\sin^2A$

• $\sin(2A)=2\sin A\cos A$

• $\sin^2A+\cos^2A=1$

Thus,

$\longrightarrow \dfrac{\cos^2\left(\dfrac{x}{2}\right)-\sin^2\left(\dfrac{x}{2}\right)}{\cos^2\left(\dfrac{x}{2}\right)+\sin^2\left(\dfrac{x}{2}\right)+2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)}=\tan y$

$\longrightarrow \dfrac{\left(\cos\left(\dfrac{x}{2}\right)-\sin\left(\dfrac{x}{2}\right)\right)\left(\cos\left(\dfrac{x}{2}\right)+\sin\left(\dfrac{x}{2}\right)\right)}{\left(\cos\left(\dfrac{x}{2}\right)+\sin\left(\dfrac{x}{2}\right)\right)^2}=\tan y$

$\longrightarrow \dfrac{\cos\left(\dfrac{x}{2}\right)-\sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right)+\sin\left(\dfrac{x}{2}\right)}=\tan y$

Dividing both numerator and denominator by $\cos\left(\dfrac{x}{2}\right),$

$\longrightarrow \dfrac{1-\tan\left(\dfrac{x}{2}\right)}{1+\tan\left(\dfrac{x}{2}\right)}=\tan y$

$\longrightarrow \dfrac{\tan\left(\dfrac{\pi}{4}\right)-\tan\left(\dfrac{x}{2}\right)}{1+\tan\left(\dfrac{\pi}{4}\right)\cdot\tan\left(\dfrac{x}{2}\right)}=\tan y$

Since $\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B},$

$\longrightarrow \tan\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=\tan y$

$\longrightarrow y=\dfrac{\pi}{4}-\dfrac{x}{2}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=-\dfrac{1}{2}}}$

Thus differentiated!

Answer 2

Given , the function is

• $\tt y = {tan}^{ - 1} \{ \frac{cos(x)}{1 + sin(x)} \}$

Differentiating y wrt x , we get

$\tt \implies \tt \frac{dy}{dx} = \frac{d \bigg \{ {tan}^{ - 1} \{ \frac{cos(x)}{1 + sin(x)} \} \bigg\}}{dx}$

$\tt \implies \frac{dy}{dx} \frac{1}{1 + { \{ \frac{cos(x)}{1 + sin(x)}\}}^{2} } \frac{d \{\frac{cos(x)}{1 + sin(x)} \}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{ { \{1 + sin(x) \} }^{2} }{ { \{1 + sin(x) \}}^{2} + {cos}^{2} (x)} \bigg\{ \frac{1 + sin(x) \frac{d \{ cos(x)\}}{dx} - cos(x) \frac{d \{1 + sin(x) \}}{dx} }{ { \{(1 + sin(x) \}}^{2} } \bigg\}$

$\tt \implies \frac{dy}{dx} = \frac{ \{1 + sin(x) \} \{- sin(x) \} - cos(x)cos(x)}{ { \{1 + sin(x) \}}^{2} + {cos}^{2}(x) }$

$\tt \implies \frac{dy}{dx} = \frac{ - sin(x) - {sin}^{2}(x) - {cos}^{2}( x) }{ {1 + {sin}^{2} (x) + 2sin(x) + {cos}^{2} (x)}}$

$\tt \implies \frac{dy}{dx} = \frac{ - sin(x) - \{ {sin}^{2} (x) + {cos}^{2}(x) \}}{1 + 2sin(x) + 1}$

$\tt \implies \frac{dy}{dx} = \frac{ - sin(x) - 1}{2 + 2sin(x)}$

$\tt \implies \frac{dy}{dx} = \frac{ - \{ sin(x) + 1\}}{2 \{1 + sin(x) \}}$

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

Remmember :

$\tt \implies \frac{d \{ {tan}^{ - 1}x\}}{dx} = \frac{1 }{1 + {(x)}^{2} }$

$\tt \implies \frac{d( \frac{u}{v} )}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {(v)}^{2} }$

$\tt \implies \frac{d \{cos(x) \}}{dx} = - sin(x)$

$\tt \implies \frac{d \{ constant\}}{dx} = 0$

$\tt \implies \frac{d \{sin(x) \}}{dx} = cos(x)$

$\tt \implies {sin}^{2} (x) + {cos}^{2} (x) = 1$