If 3(\sec ^{2} \theta+\tan ^{2} \theta)=5,0^{\circ}

SSC CHSL 16 Apr 2021 Shift 2 Paper

Section: Quantitative Aptitude

Question : 53 of 100

Marks: +1, -0

3(\sec ^{2} θ+\tan ^{2} θ)=5,0^{∘} < θ < 90^{∘}

, then the value of

c o s e c θ

Solution:

Given,

⇒\;3(\sec ^{2} θ+\tan ^{2} θ)=5

⇒\;3(\tan ^{2} θ+1+\tan ^{2} θ)=5

\;3+6 \tan ^{2} θ=5

⇒\;\tan ^{2} θ=\;{2}/{6}=\;{1}/{3}

⇒\;\tan θ=\;{1}/{√{3}}=\tan 30^{∘}

∴ \;{[ ∵ 0^{∘}< θ< 90^{∘}]}

∴ \;cosec θ=cosec 30^{∘}=2

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