Question

prove that following identity

cosA/(1-tanA) + sinA/(1-cotA) = cosA + sinA
​

Answer 1

$\huge \bold{ \fbox{ \underline \pink{required \: answer}}}$

we have,

$\bold{LHS = \frac{ \cos A}{1 - \tan A} + \frac{ \sin }{1 - \cot A \: } } \\ \\ \bold{RHS = \cos A + \sin A} \:$

now solve LHS;

${ \tt\bold{LHS = \frac{ \cos A}{1 - \tan A} + \frac{ \sin }{1 - \cot A \: }} } \\ \\ \bold{ { \tt \implies LHS = \frac{ \cos A}{1 - \frac{ \sin A}{ \cos A} } \: + \: \frac{ \sin A}{1 - \frac{ \cos A}{ \sin A} } }} \\ \\ \bold{ \tt \implies LHS = \frac{ \cos A}{ \frac{ \cos A - \sin A}{ \cos A} } \: + \: \frac{ \sin A}{ \frac{ \sin A - \cos A}{ \sin A} } } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2}A}{ \cos A - \sin A} \: + \: \frac{ \sin^{2}A }{ \sin A - \cos A} } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2} A}{ \cos A - \sin A} \: - \: \frac{ \sin^{2}A }{ \cos A - \sin A} } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2}A - \sin^{2}A }{ \cos A - \sin A} } \\ \\ \orange{\bold{ \tt \: since ,\: \: {a}^{2} - {b}^{2} = (a - b)(a + b) } } \\ \\ \bold{ \tt \implies LHS = \frac{( \cos A - \sin A)( \cos A + \sin A)}{ \cos A - \sin A} } \\ \\ \bold{ \tt \implies LHS = \cos A + \sin A \: = \: RHS }$

so, LHS = RHS

${ \bold{{ \underline{ \underline \red{hence, \: proved}}}}}$

Answer 2

Answer:

$\huge \bold{ \fbox{ \underline \pink{required \: answer}}} required answer</p><p></p><p>we have,</p><p></p><p>\begin{gathered} \bold{LHS = \frac{ \cos A}{1 - \tan A} + \frac{ \sin }{1 - \cot A \: } } \\ \\ \bold{RHS = \cos A + \sin A} \: \end{gathered}LHS=1−tanAcosA+1−cotAsinRHS=cosA+sinA</p><p></p><p>now solve LHS;</p><p></p><p>\begin{gathered}{ \tt\bold{LHS = \frac{ \cos A}{1 - \tan A} + \frac{ \sin }{1 - \cot A \: }} } \\ \\ \bold{ { \tt \implies LHS = \frac{ \cos A}{1 - \frac{ \sin A}{ \cos A} } \: + \: \frac{ \sin A}{1 - \frac{ \cos A}{ \sin A} } }} \\ \\ \bold{ \tt \implies LHS = \frac{ \cos A}{ \frac{ \cos A - \sin A}{ \cos A} } \: + \: \frac{ \sin A}{ \frac{ \sin A - \cos A}{ \sin A} } } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2}A}{ \cos A - \sin A} \: + \: \frac{ \sin^{2}A }{ \sin A - \cos A} } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2} A}{ \cos A - \sin A} \: - \: \frac{ \sin^{2}A }{ \cos A - \sin A} } \\ \\ \bold{ \tt \implies LHS = \frac{ \cos^{2}A - \sin^{2}A }{ \cos A - \sin A} } \\ \\ \orange{\bold{ \tt \: since ,\: \: {a}^{2} - {b}^{2} = (a - b)(a + b) } } \\ \\ \bold{ \tt \implies LHS = \frac{( \cos A - \sin A)( \cos A + \sin A)}{ \cos A - \sin A} } \\ \\ \bold{ \tt \implies LHS = \cos A + \sin A \: = \: RHS }\end{gathered}LHS=1−tanAcosA+1−cotAsin⟹LHS=1−cosAsinAcosA+1−sinAcosAsinA⟹LHS=cosAcosA−sinAcosA+sinAsinA−cosAsinA⟹LHS=cosA−sinAcos2A+sinA−cosAsin2A⟹LHS=cosA−sinAcos2A−cosA−sinAsin2A⟹LHS=cosA−sinAcos2A−sin2Asince,a2−b2=(a−b)(a+b)⟹LHS=cosA−sinA(cosA−sinA)(cosA+sinA)⟹LHS=cosA+sinA=RHS</p><p></p><p>so, LHS = RHS</p><p></p><p>{ \bold{{ \underline{ \underline \red{hence, \: proved}}}}}hence,proved</p><p></p><p>$