can anyone pl explain me how this step came
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Those who are common, have been written together
(2-2sinA) (1+cos A)
(2-2sinA) (1+cos A)
Anonymous:
thanku
Answered by
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Answer:
[1 - sinA + cosA]² = 2(1 - sinA)(1 + cosA)
Step-by-step explanation:
LHS:
= [1 - sinA + cosA]²
= (1)² + (-sinA)² + (cosA)² + 2(1)(-sinA) + 2(-sinA)(cosA) + (2cosA)(1)
= 1 + sin²A + cos²A - 2 sinA - 2 sinA. cosA + 2 cosA
= 1 + 1 + 2 cosA - 2 sinA - 2sinA.cosA
= 2 + 2 cosA - 2 sinA - 2 sinA . cosA
= 2 - 2 sinA + 2 cosA - 2 sinA.cosA
= 2(1 - sinA) + 2 cosA(1 - sinA)
= (2 + 2 cosA)(1 - sinA)
= 2[1 + cosA][1 - sinA]
= RHS
Hope it helps!
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