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CBSE Class 10 Answered

evaluate the following
Asked by sunil_0010 | 31 Oct, 2009, 08:46: PM
answered-by-expert Expert Answer

[(1+sinθ-cosθ)/(1+sinθ+cosθ)]2 =

(1+sinθ-cosθ)2/(1+sinθ+cosθ)2 =

(1+2sinθ+sin2θ-2cosθ-2sinθcosθ+cos2θ)/(1+2sinθ+sin2θ-2cosθ-2sinθcosθ+cos2θ) =

(1+2sinθ+1-2cosθ-2sinθcosθ)/(1+2sinθ+1+2cosθ+2sinθcosθ) = 

(2+2sinθ-2cosθ-2sinθcosθ)/(2+2sinθ+2cosθ+2sinθcosθ) =

(1+sinθ-cosθ-sinθcosθ)/(1+sinθ+cosθ+sinθcosθ) =

((1+sinθ)-cosθ(1+sinθ))/((1+sinθ)+cosθ(1+sinθ)) =

(1+sinθ)(1-cosθ)/((1+sinθ)(1+cosθ)) =

(1-cosθ)/(1+cosθ)

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Answered by | 01 Nov, 2009, 11:20: PM

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