Question

1/secΦ+tanΦ = 1-sinΦ/cosΦ​

Answer 1

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.We have to prove that,sinθ−cosθ+1sinθ+cosθ−1=1secθ−tanθ using identity sec2θ=1+tan2θ

LHS = sinθ−cosθ+1sinθ+cosθ−1=tanθ−1+secθtanθ+1−secθ [ dividing the numerator and denominator by cosθ.]

=(tanθ+secθ)−1(tanθ−secθ)+1={(tanθ+secθ)−1}(tanθ−secθ){(tanθ−secθ)+1}(tanθ−secθ) [ Multiplying and dividing by (tanθ−secθ)]

=(tan2θ−sec2θ)−(tanθ−secθ){(tanθ−secθ)+1}(tanθ−secθ) [∵(a−b)(a+b)=a2−b2]

=−1−tanθ+secθ(tanθ−secθ+1)(tanθ−secθ)[∵tan2θ−sec2θ=−1]

=−(tanθ−secθ+1)(tanθ−secθ+1)(tanθ−secθ)=−1tanθ−secθ

=1secθ−tanθ=RHS

Hence Proved.

Answer 2

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