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SSC Selection Post 2022 PYQ — Question 68

SSC Selection Post - Higher Secondary Level - 09 Feb 2022 - Shift 2 • Quantitative Aptitude • Q68

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If 3 tan θ = 23{2 \sqrt{3} } sin θ, 0° < θ < 90°, then the value of 34(cosec22θ+cot22θsin2θ+tan22θ){3\over 4} ({cosec^2 2\theta + cot^2 2\theta {} \over sin^2 \theta + tan^2 2\theta}) is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. 513{5\over 13} . 

3 tan θ = 23{2 \sqrt{3} } sin θ

3 sinθ/cosθ = 23{2 \sqrt{3} } sin θ

3cosθ3\over cos \theta = 23{2 \sqrt{3} }

θ = 30°

⇒ 34(cosec22θ+cot22θsin2θ+tan22θ){3\over 4} ({cosec^2 2\theta + cot^2 2\theta {} \over sin^2 \theta + tan^2 2\theta})

⇒ 34(cosec260+cot260sin230+tan260){3\over 4} ({cosec^2 60 + cot^2 60 {} \over sin^2 30 + tan^2 60})

⇒ 34((23)2+(13)2(12)2+(3)2){3\over 4} ({({2\over\sqrt3})^2 + ({1\over\sqrt3})^2 \over ({1\over2})^2 +({\sqrt3})^2 })

⇒ 34(43+1314+3){3\over 4} ({{4\over3} + {1\over3} \over {1\over4} +3 })

⇒ 34(53134){3\over 4} ({{5\over3} \over {13\over4} })

⇒ 34×53×413{3\over 4}\times {{5\over3} \times {4\over13} } = 513{5\over 13}