Question

please solve this
continue with what I have done

Answer 1

[0010001111]... Hello User... [1001010101]
Here's your answer...

In this sum, we'll use the identity
(a+b+c)² = a²+b²+c²+2(ab+bc+ca)

${( \frac{1 + \sin \alpha - \cos \alpha }{1 + \sin \alpha + \cos \alpha } )}^{2} \\ = \frac{1 + { \sin \alpha }^{2} + { \cos }^{2} \alpha + 2( \sin \alpha - \cos \alpha - \sin \alpha \cos \alpha ) }{1 + { \sin}^{2} \alpha + { \cos }^{2} \alpha + 2( \sin \alpha + \cos \alpha + \sin \alpha \cos \alpha ) } \\ = \frac{2(1 + \sin \alpha - \cos\alpha - \sin \alpha \cos \alpha ) }{2(1 + \sin \alpha + \cos \alpha + \sin \alpha \cos \alpha ) } \\ = \frac{(1 + \sin \alpha ) - \cos \alpha (1 + \sin \alpha )}{(1 + \sin \alpha ) + \cos \alpha (1 + \sin \alpha ) } \\ = \frac{(1 + \sin\alpha )(1 - \cos \alpha ) }{(1 + \sin \alpha )(1 + \cos \alpha ) } \\ = \frac{1 - \cos \alpha }{1 + \cos \alpha }$

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