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SSC CHSL 2021 PYQ — Question 70

SSC CHSL - 6 Aug 2021 - Shift 3 • Quantitative Aptitude • Q70

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If sin2θtan2θ−sin2θ=5\frac{sin^2\theta}{tan^2\theta-sin^2\theta}=5, then the value of 24cos2θ−15sec2θ6cosec2θ−7cot2θ\frac{24cos^2\theta-15sec^2\theta}{6cosec^2\theta-7cot^2\theta} is:

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Correct answer: C

Explanation

The correct answer is option 3 i.e. 2

Given:

sin2θtan2θ−sin2θ=5\frac{sin^2\theta}{tan^2\theta-sin^2\theta}=5

Concept USed:

cos A = BHB \over H

Explanation:

According to the given question,

sin2θtan2θ−sin2θ=5\frac{sin^2\theta}{tan^2\theta-sin^2\theta}=5

⇒ sin2θ=5(tan2θ−sin2θ){sin^2\theta} = 5({tan^2\theta-sin^2\theta})

⇒ 6sin2θ=5(sin2θcos2θ)6{sin^2\theta} = 5( {sin^2\theta \over cos^2 \theta})

⇒ cos⁡2θ=56\cos^2 \theta = {5 \over 6}

⇒ cos⁡θ=56\cos \theta = \sqrt{5 \over 6}

Now, sec⁡θ=65; cosecθ=6; cotθ=5\sec \theta = \sqrt{6 \over 5};\ cosec \theta = \sqrt{6};\ cot\theta = \sqrt{5}

Putting the values, we get

24×56−15×656×6−7×5\frac{24 \times {5 \over 6}-15 \times{6 \over 5}}{6\times 6-7\times 5} = 20−1836−35\frac{20- 18}{36-35} = 2

Hence the required answer is 2.