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(1 + sin theta)^2 + cos^2 theta = 2(1 + sin theta) |
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Answer» Taking LHS= 2 θ (1+sinθ) 2 (1−sinθ) 2
= 2cos 2 θ 1+sin 2 θ+2sinθ+1+sin 2 θ−2sinθ
[(a+B) 2 =a 2 +b 2 +2AB;(a−b) 2 =a 2 +b 2 −2ab] = 2cos 2 θ 2+2sin 2 θ
= 2(1−sin 2 θ) 2(1+sin 2 θ)
[∵cos 2 θ+sin 2 θ=1] = (1−sin 2 θ) (1+sin 2 θ)
=RHS |
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