prove that : (1+sinΘ + cosΘ)^2 = 2 (1+sinΘ) (1+cosΘ)
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[(1+sinθ)+cosθ]^2
= (1+sinθ)^2+2(1+sinθ)cosθ+cos^2θ
= 1+2sinθ+sin^2θ+2cosθ+2sinθcosθ+cos^2θ
= 2+2(sinθ+cosθ)+2sinθcosθ {sin^2θ+cos^2θ = 1}
= 2(1+sinθ)+2cosθ(1+sinθ)
= (1+sinθ)(2+2cosθ)
= 2(1+cosθ)(1+sinθ)
L.H.S= R.H.S
[(1+sinθ)+cosθ]^2
= (1+sinθ)^2+2(1+sinθ)cosθ+cos^2θ
= 1+2sinθ+sin^2θ+2cosθ+2sinθcosθ+cos^2θ
= 2+2(sinθ+cosθ)+2sinθcosθ {sin^2θ+cos^2θ = 1}
= 2(1+sinθ)+2cosθ(1+sinθ)
= (1+sinθ)(2+2cosθ)
= 2(1+cosθ)(1+sinθ)
L.H.S= R.H.S
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