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SSC Selection Post 2022 PYQ — Question 66

SSC Selection Post - Higher Secondary Level 2022 - 15 March 2022 - Shift 2 • Quantitative Aptitude • Q66

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If cos2θ−sin2θ−3cosθ+2=0,00<θ<900,{{cos^2\theta-sin^2\theta-3cos\theta+2=0,0⁰<\theta<90⁰,}} then what is the value of 1+5cosθ+tanθ2?1 +5cos\theta+{{tan\theta} \over 2}?

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Correct answer: B

Explanation

The correct answer is Option 2 i.e. 7+32{7+ \sqrt{3} \over 2}

cos⁡2θ−sin⁡2θ−3cos⁡θ+2=0\cos^2\theta - \sin^2\theta - 3\cos\theta + 2 = 0

⇒2cos⁡2θ−3cos⁡θ+1=0\Rightarrow 2\cos^2\theta - 3\cos\theta + 1 = 0

cos⁡θ=12⇒θ=60∘\cos\theta = \frac{1}{2} \Rightarrow \theta = 60^\circ

1+5cos⁡θ+tan⁡θ2=1+52+32=7+32 1 + 5\cos\theta + \frac{\tan\theta}{2} = 1 + \frac{5}{2} + \frac{\sqrt{3}}{2} = \frac{7 + \sqrt{3}}{2}