Question

express tan^-1(cosx+sinx/cosx-sinx)in the simplest form
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Answer 1

$\implies\bf{ \frac{\pi}{4} - x }$

Explanation

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

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

$\implies\sf{ \frac{ \cancel{\frac{ \cos \: x }{ \cos \: x}} - \frac{ \sin \: x }{ \cos \: x } }{ \cancel{\frac{ \cos \: x}{ \cos \: x}} + \frac{ \sin \: x }{ \cos \: x} }}$

$\implies\sf{ { \tan }^{ - 1}( \frac{1 - \tan \: x }{1 + \tan \: x }) }$

$\implies\sf{ { \tan }^{ - 1}( \frac{1 - \tan \: x}{1 + 1 \times \tan \: x }) }$

$\implies\sf{ { \tan }^{ - 1}( \frac{ \tan \frac{\pi}{4} - \tan \: x }{1 + \tan\frac{\pi}{4} \times \tan \: x }) }$

$\implies\sf{\cancel{{ \tan }^{ - 1}\{\tan}( \frac{\pi}{4} - x)\}}$

$\implies\boxed{\bf{ \frac{\pi}{4} - x }}$

Identities used

$\diamond\boxed{\bf{\frac{\cos\:x-\sin\:x}{\cos\:x+\sin\:x}=\tan}}$$\diamond\boxed{\bf{tan(x-y)\frac{\tan\:x-\tan\:y}{1+\tan\:x\tan\:y}}}$$\diamond\boxed{\bf{\tan\frac{\pi}{4}=1}}$

Hope it helps!!