Which of the following options is equal to sin θ+1−cos θcos θ−1+sin θ?
tanθ+secθ
sinθ+1−cosθcosθ−1+sinθ
Dividing the numerator and denominator by cos θ we get,
sinθcosθ+1cos θ−cosθcosθcosθcosθ−1cosθ+sinθcosθ
=tanθ+secθ−11−secθ+tanθ
⇒(tanθ+secθ)−(sec2θ−tan2θ)1−secθ+tanθ....(1+tan2θ=sec2θ)
=(tanθ+secθ)−(secθ−tanθ)(secθ+tanθ)1−secθ+tanθ
=(tanθ+secθ)(1−secθ+tanθ)(1−secθ+tanθ)
=tanθ+secθ
=sinθcosθ+1cosθ
=sinθ+1cosθ
[Multiplying the numerator and denominator by (1−sinθ)]
=1+sinθcosθ×1−sinθ1−sinθ
=1−sin2θcosθ(1−sinθ)
=cos2θcosθ(1−sinθ)
=cosθ1−sinθ