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SSC CGL Tier II - Quant - 15 Nov 2020

Tier 2 • 15 November 2020 • 100 questions

Source label: Previous Paper

Question 1 / 100

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

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All 100 questions in this paper
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  31. Q31.If sec = a/b , b ≠ 0, then (1 - tan²θ) / ( 2 - Sin²θ) = ?
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