Question

prove that using the identity
sec ²θ = 1 + tan² θ

sin θ - cos θ + 1 / sin θ + cos θ - 1
== 1 / sec θ - tan θ




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Answer 1

Answer:

=sinθ−cosθ+1sinθ+cosθ−1

=sinθcosθ−1+1cosθsinθcosθ+1−1cosθ

[ on dividing num. and denom. by cosθ ]

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

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

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

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

RHS=1(secθ−tanθ)

=1(secθ−tanθ)×(secθ+tanθ)(secθ+tanθ)=(secθ+tanθ)(sec2θ−tan2θ)

=(secθ+tanθ) [∵sec2θ−tan2θ=1].

Hence,LHS=RHS.

Explanation:

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