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SSC CGL 2021 PYQ — Question 71

SSC CGL - 17 Aug 2021 - Shift 2 • Quantitative Aptitude • Q71

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For 0° < θ < 90°, 1cosθ+1tanθ−secθ\frac{1}{cos θ}+\frac{1}{tan θ - sec θ} is equal to:

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Correct answer: C

Explanation

The correct answer is Option 3 i.e - tan θ.

⇒1cosθ+1tanθ−secθ\frac{1}{cos θ}+\frac{1}{tan θ - sec θ}

⇒1cosθ+1sinθcosθ−1cosθ\frac{1}{cos θ}+\frac{1}{\frac{sinθ}{cosθ} - \frac{1}{cosθ}}

 ⇒1cosθ+cosθsinθ−1\frac{1}{cos θ}+\frac{cosθ}{sinθ - 1}

  ⇒sinθ−1+cos2θcosθ×(sinθ−1)\frac{sinθ - 1 + cos^2 θ}{cosθ \times (sinθ - 1)}

⇒sinθ−sin2θcosθ×(sinθ−1)\frac{sinθ - sin^2 θ}{cosθ \times (sinθ - 1)}

⇒sinθ−cosθ=−tanθ\frac{sin \theta}{- cos \theta} = -tan\theta