Question

please solve this its urgent Please give the answer as an attachment​

Answer 1

Answer:

here u go

Step-by-step explanation:

follow this attachment..hope u understand

Answer 2

Step-by-step explanation:

(tan A /1-tan A)-( cot A/1-cot A )= (cos A+sin A)/(cos A-sin A)

LHS=(tan A /1-tan A)-( cot A/1-cot A )

$= \frac{ \frac{ \sin \: \alpha }{ \cos( \alpha ) } }{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } - \frac{ \frac{ \cos( \alpha ) }{ \ \sin( \alpha ) } }{1 - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } } \\ = \frac{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{ \frac{ \cos( \alpha ) - \sin( \alpha ) }{ \cos( \alpha ) } } - \frac{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } }{ \frac{ \sin( \alpha ) - \cos( \alpha ) }{ \sin( \alpha ) } } \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{ \cos( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \times \frac{ \sin( \alpha ) }{ \sin( \alpha ) - \cos( \alpha ) } \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } - \frac{ \cos( \alpha ) }{ \sin( \alpha ) - \cos( \alpha ) } \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } - \frac{ \cos( \alpha ) }{ - (\cos( \alpha ) - \sin( \alpha )) } \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } + \frac{ \cos( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } \\ = \frac{ \sin( \alpha ) + \cos( \alpha ) }{ \cos( \alpha) - \sin( \alpha ) }$