Question

prove that Cos A by 1 minus tan A + sin a by 1 minus cot a =
Sin A + Cos A​

Answer 1

Step-by-step explanation:

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Answer 2

Hey Pretty Stranger!

$\sf \: LHS = \dfrac{ \cos \: A }{1 - \tan \: A } + \dfrac{ \sin \:A }{1 - \cot \: A}$

$\sf \hookrightarrow \: \dfrac{ \cos \:A }{1 - ( \frac{ \sin \: A }{ \cos \:A } )} + \dfrac{ \sin \:A }{1 - ( \frac{ \cos \: A }{ \sin \:A } )}$

$\sf \hookrightarrow \: \dfrac{ { \cos}^{2} A}{ \cos \: A \: - \sin \: A} + \dfrac{ { \sin}^{2} A}{ \cos \: A \: - \sin \: A}$

$\sf \hookrightarrow \: \dfrac{ { \cos}^{2} A}{ \cos \: A \: - \sin \: A} - \dfrac{ { \sin}^{2} A}{ \cos \: A \: - \sin \: A}$

$\sf \hookrightarrow \: \dfrac{ { \cos}^{2} A - { \sin}^{2}A }{ \cos \: A - \sin \: A}$

$\sf \hookrightarrow \: \dfrac{ (\cos \: A - \sin \: A) ( \cos \:A + \sin \: A }{ (\cos \: A - \sin \: A) }$

$\sf \hookrightarrow \: (\cos \: A + \sin \: A)$

$\sf \hookrightarrow \: ( \sin \: A + \cos \: A \: )$

Hence,Proved!