Math, asked by suriesiva92, 1 month ago

prove that sec^2θ sin^2θ-cosec^2θ+cosec^2θcos^2θ/sec^2θsin^2θ-cosec^2θcos^2θ=sin^2θ​

Answers

Answered by sudhirsingh0779
3

Answer:

L.H.S. = sec2 θ + cosec2 θ = 1 c o s 2 θ + 1 s i n 2 θ 1cos2θ+1sin2θ = s i n 2 θ + c o s 2 θ c o s 2 θ . s i n 2 θ = 1 c o s 2 θ . s i n 2 θ sin2θ+cos2θcos2θ.sin2θ=1cos2θ.sin2θ (∵ sin2 θ + cos2 θ = 1) = 1 c o s 2 θ 1 s i n 2 θ 1cos2θ1sin2θ = sec2 θ

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