Question

cosA/1-tanA +sinA/1-cotA=sinA+cosA​

Answer 1

Answer is in attachment

Answer 2

Step-by-step explanation:

$\frac{cos \alpha }{1 - tan \alpha } + \frac{sin \alpha }{1 - cot \alpha } \\ \\ = \frac{cos \alpha }{1 - tan \alpha } + \frac{sin \alpha }{1 - \frac{1}{tan \alpha } } \\ \\ = \frac{cos \alpha }{1 - tan \alpha } + \frac{sin \alpha }{ \frac{tan \alpha - 1}{tan \alpha } } \\ \\ = \frac{cos \alpha }{1 - tan \alpha } + \frac{tan \alpha \: sin \alpha }{tan \alpha - 1} \\ \\ = \frac{cos \alpha }{1 - tan \alpha } - \frac{tan \alpha \: sin \alpha }{1 - tan \alpha } \\ \\ = \frac{cos \alpha - tan \alpha \: \: sin \alpha}{1 - tan \alpha } \\ \\ = \frac{cos \alpha - \frac{sin \alpha }{cos \alpha }sin \alpha }{1 - \frac{sin \alpha }{cos \alpha } } \\ \\ = \frac{ {cos}^{2} \alpha - {sin}^{2} \alpha }{cos \alpha - sin \alpha } \\ \\ = \frac{(cos \alpha - sin \alpha )(cos \alpha + sin \alpha )}{(cos \alpha - sin \alpha )} \\ \\ = cos \alpha + sin \alpha \: \: \: \: \: \: \: \: (hence \: proved)$

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