SarkariPYQ

SSC CGL 2024 PYQ — Question 59

SSC CGL - 12 Sep 2024 - Shift 1 • Quantitative Aptitude • Q59

Back to paper

The value of (sinθ−2sin3θ2cos3θ−cosθ)31tanθ−sec2θ{\left(sin\theta-2sin^3\theta\over 2cos^3\theta-cos\theta\right)}^3 {1\over tan\theta} {-sec^2\theta} is:

Show answer & explanation

Correct answer: D

Explanation

The correct answer is Option 4 i.e. −1.

⇒ (sinθ−2sin3θ2cos3θ−cosθ)3×1tanθ−sec2θ({sin\theta-2sin^3\theta\over 2cos^3\theta-cos\theta})^3\times {1\over tan\theta} {-sec^2\theta}

⇒ (sinθ(1−2sin2θ)cosθ(2cos2θ−1))3×1tanθ−sec2θ({sin\theta (1-2sin^2\theta)\over cos\theta(2cos^2\theta-1)})^3\times {1\over tan\theta} {-sec^2\theta}

[1 - 2sin2θ = cos2θ, 2cos2θ - 1 = cos2θ]

⇒ (sinθ×cos2​​​​​​θcosθ×cos2θ)3×1tanθ−sec2θ({sin\theta \times cos2​​​​​​\theta \over cos\theta \times cos2\theta})^3\times {1\over tan\theta} {-sec^2\theta}

⇒ (sinθcosθ)3×1tanθ−sec2θ({sin\theta\over cos\theta})^3\times {1\over tan\theta} {-sec^2\theta}

⇒ tan2θ−sec2θ{tan^2\theta}{-sec^2\theta} 

[1 + tan2θ = sec2θ]

⇒ −(sec2θ−tan2θ)-({sec^2\theta}{-tan^2\theta}) = -1