sinθ[(1−tanθ)tanθ+sec2θ]/(1−sinθ)tanθ(1+tanθ)(secθ+tanθ) is equal to:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 1
GIVEN:
sinθ[(1−tanθ)tanθ+sec2θ]/(1−sinθ)tanθ(1+tanθ)(secθ+tanθ)
We know:
sec2θ−tan2θ=1, sin2θ + cos2θ = 1, tanθ = sinθ/cosθ, secθ = 1/cosθ
Concept
= sinθ[(1−tanθ)tanθ+sec2θ]/(1−sinθ)tanθ(1+tanθ)(secθ+tanθ)
=sinθ[tanθ−tan2θ+1+tan2θ]/(1−sinθ)tanθ(1+tanθ)(secθ+tanθ)
=sinθ(1+tanθ)/(1−sinθ)tanθ(1+tanθ)(secθ+tanθ)
=sinθ/(1−sinθ)sinθ/cosθ(secθ+tanθ)
=cosθ/(1−sinθ)(1/cosθ+sinθ/cosθ)
=cos2θ/(1−sinθ)(1+sinθ)
=cos2θ/(1−sin2θ)
=cos2θ/cos2θ
= 1