If \sec (90^{\circ}-\frac{3 \theta}{2})=\sqrt{2}, 0^{\circ}

SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 56 of 100

Marks: +1, -0

\sec (90^{∘}-\;{3 θ}/{2})=√{2}, 0^{∘}< θ< 90^{∘}

, then the value of

2 \sin θ+4 \cos 2 θ

will be

Solution:

Given,

\sec (90^{∘}-\;{3 θ}/{2})=√{2}

⇒ cosec \;{3 θ}/{2}=√{2}=cosec 45^{∘}

On comparison,

\;\; [ ∵ 0^{∘} < θ < 90^{∘}]

⇒\;\;{3 θ}/{2}=45^{∘}

⇒\;θ=\;{45^{∘}}/{3} × 2=30^{∘}

∴

The value of

2 \sin θ+4 \cos 2 θ

\;\; =2 \sin 30^{∘}+4 \cos (2 × 30^{∘})

\;\; =2 \sin 30^{∘}+4 \cos 60^{∘}

\;\; =2 × \;{1}/{2}+4 × \;{1}/{2}=1+2=3

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