Question

differentiate with respect to x :-

tan inverse (sinx/1+cosx)

Answer 1

$\textbf{Hey there}$!


$\textbf{Solution}$:


$\huge \: y$= $\huge\tan ^{ - 1} (\frac{ sinx}{1 + cosx})$


$\huge{put \: x \: = 2 \theta}$


$\huge \theta \: = \frac{x}{2} .......(1)$


$\huge \: y$= $\huge\tan ^{ - 1} (\frac{ sin2 \theta}{1 + cos2 \theta})$


Using Identities:


sin2x = 2sinxcosx.


1 + cos2x = 2$cos^{2}$x.


$\huge \: y$=$\huge\tan ^{ - 1} (\frac{ 2sin \theta \: cos \theta}{2cos^{2} \theta })$


$\huge \: y$= $\huge\tan ^{ - 1} (\frac{ 2 \: sin \theta \: \cancel {cos \theta}}{2cos^{ \cancel2} \theta })$


$\huge \: y$= $\huge\tan ^{ - 1} (\frac{ \cancel2sin \theta }{ \cancel2cos\theta })$


$\huge \: y$= $\huge\tan ^{ - 1} (\frac{ sin \theta}{cos \theta })$


$\huge \: y$= $\huge\tan ^{ - 1} (tan \theta)$


$\huge \: y$= $\huge \cancel tan ^{ - 1} ( \cancel tan \theta)$


$\huge \: y$= $\huge\theta$


Using Equation ( 1 ).


Therefore,


$\huge \ y = \frac{x}{2}$


Now , Differentiate Both Sides w.r.t x


we get:


$\huge \: dy/dx$=$\huge \ \frac{1}{2}$



#Be Brainly.

Answer 2

The given equation is:

$tan^{-1}(\frac{sinx}{1+cosx})$

Upon solving the above equation, we get

$=tan^{-1}(\frac{2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}})$

$=tan^{-1}(\frac{sin\frac{x}{2}}{cos\frac{x}{2}})$

$=tan^{-1}(tan\frac{x}{2})$

$=\frac{x}{2}$

Which is the required simplified form.