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SSC CGL 2021 PYQ — Question 70

SSC CGL - 23 Aug 2021 - Shift 1 • Quantitative Aptitude • Q70

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if sin2 θ - cos2 θ - 3sinθ + 2 = 0, 0° < θ < 90°, then what is the value of 1secθ−tan⁡θ\frac{1}{\sqrt{sec\theta-\tan\theta}} is:

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Correct answer: A

Explanation

The correct answer is Option 1 i.e  34\sqrt[4]{3}.

sin2 θ - cos2 θ - 3sinθ + 2 = 0

Let's assume the value of θ = 30°

sin2 30° - cos2 30° - 3sin30° + 2 = 0

⇒ (1/4) - (3/4) - (3/2) + 2 = 0

⇒ - 2 + 2 =0

⇒ 0 = 0

LHS = RHS 

We have to find the value of 1secθ−tan⁡θ\dfrac{1}{\sqrt{sec\theta-\tan\theta}}

Put  θ = 30° 

1sec30−tan30\dfrac{1}{\sqrt{sec30-tan30}}

⇒ 123−13\dfrac{1}{\sqrt{\frac{2}{\sqrt3}-\frac{1}{\sqrt3}}}

⇒ 113\dfrac{1}{\sqrt\frac{1}{\sqrt3}}

⇒ 3\sqrt{\sqrt{3}}

⇒ 34\sqrt[4]{3}