Question

2nd part solution fast plz​

Answer 1

Answer: LHS :

2sec2 θ−sec 4 θ−2cosec 2 θ+cosec 4

=(cosec 4 θ−2cosec 2 θ)−(sec 4 θ−2sec 2 θ)

=(cosec 4 θ−2cosec 2 θ+1)−(sec 4 θ−2sec 2 θ+1)

=(cosec 2 θ−1) 2 −(sec 2 θ−1) 2

=cot 4 θ−tan 4 θ

= RHS

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Answer 2

Answer:

Please refer to the attached image to clarify your doubt..

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