Question

differentiate tan^(-1)(secx+tanx)w.r.to x is

Answer 1

Step-by-step explanation:

We have,

$y = \tan^{ - 1} \bigg( \sec(x) + \tan(x) \bigg)$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{ 1+ \sin(x) }{ \cos(x) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{ 1+ 2\sin( \frac{x}{2} ) \cos( \frac{x}{2} ) }{ \cos^{2} ( \frac{x}{2} ) - \sin^{2} ( \frac{x}{2} ) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{\cos^{2} ( \frac{x}{2} ) + \sin^{2} ( \frac{x}{2} ) + 2\sin( \frac{x}{2} ) \cos( \frac{x}{2} ) }{ \cos^{2} ( \frac{x}{2} ) - \sin^{2} ( \frac{x}{2} ) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{(\cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ))^{2} }{ (\cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ))( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{\cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ) }{ \cos ( \frac{x}{2} ) - \sin ( \frac{x}{2}) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{1 + \tan ( \frac{x}{2} ) }{ 1 - \tan ( \frac{x}{2}) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\frac{ \tan(\frac{\pi}{4} ) + \tan ( \frac{x}{2} ) }{ 1 - \tan( \frac{\pi}{4} ) \tan ( \frac{x}{2}) } \bigg \}$

$\implies \: y = \tan^{ - 1} \bigg \{\tan \bigg(\frac{\pi}{4} + \frac{x}{2} \bigg) \bigg \}$

$\implies \: y = \frac{\pi}{4} + \frac{x}{2}$

$\implies \: \frac{dy}{dx} = \frac{1}{2}$