Question

cos theta / (1-tan theta) + sin theta (1 - cot theta) =sin theta + cos theta​

Answer 1

Step-by-step explanation:

$\begin{gathered}\small{ = > \frac{ \cos A }{1 - \tan A } + \frac{ \sin A }{1 - \cot A} } \\ \\ \\ \small{ = > \frac{ \cos A }{1 - \frac{ \sin A }{ \cos A } } + \frac{ \sin A }{1 - \frac{ \cos A}{ \sin A } }} \\ \\ \\ \small{= > \frac{ \cos A }{ \frac{ \cos A - \sin A }{ \cos A} } + \frac{ \sin A }{ \frac{ \sin A - \cos A}{ \sin A } }} \\ \\ \\ \small{ = > \frac{ \cos ^{2} A }{ \cos A - \sin A} + \frac{ \sin^{2} A }{ \sin A - \cos A} } \\ \\ \\ \small{ = > \bold{ - } \frac{ \cos ^{2} A}{ \sin A - \cos A } + \frac{ \sin^{2} A }{ \sin A - \cos A} } \\ \\ \\ \small{ = > \frac{ \sin ^{2} A- \cos^{2} A }{ \sin A - \cos A }} \\ \\ \\ \small{= > \frac{( \sin A - \cos A )( \sin A + \cos A )}{ \sin A - \cos A }} \\ \\ \\ \small{ = > \sin A + \cos A} \end{gathered}=>1−tanAcosA+1−cotAsinA=>1−cosAsinAcosA+1−sinAcosAsinA=>cosAcosA−sinAcosA+sinAsinA−cosAsinA=>cosA−sinAcos2A+sinA−cosAsin2A=>−sinA−cosAcos2A+sinA−cosAsin2A=>sinA−cosAsin2A−cos2A=>sinA−cosA(sinA−cosA)(sinA+cosA)=>sinA+cosA</p><p></p><p>Hence, proved.</p><p></p><p>$