Question

(1-sinA+cosA)square=2(1+cos)(1-sinA)​

Answer 1

Answer:

Step-by-step explanation:

Consider the LHS: (1-sinA+cosA)² = [(1-sinA) + cosA]²

=  (1-sinA)² + cos2A + 2(1-sinA)cosA

=  1 + sin²A − 2sinA + cos²A + 2(1-sinA)cosA  

=  1 + (sin²A + cos²A) − 2sinA + 2(1-sinA)cosA  

=  1 + 1 − 2sinA + 2(1-sinA)cosA    [Since, sin²A + cos²A =1]

=  2 − 2sinA + 2(1-sinA)cosA

=  2(1 − sinA) + 2(1-sinA)cosA

=  2(1 − sinA)(1 + cosA)

= RHS

Answer 2

I hope it is clear to you

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