If \sec \theta-cosec \theta=0 and \theta is an acute angle, then what is the value of \sec ^{2} \theta+cosec^{2} \theta ?

SSC CHSL 12 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 70 of 99

Marks: +1, -0

\sec θ-cosec θ=0

and

θ

is an acute angle, then what is the value of

\sec ^{2} θ+cosec^{2} θ

Solution:

Given,

\sec θ-cosec θ=0

⇒ \sec θ=cosec θ

⇒ \;{1}/{\cos θ}=\;{1}/{\sin θ} ⇒ \;{\sin θ}/{\cos θ}=1

⇒ \;\; \tan θ=1

Since,

θ

is an acute angle,

∴ \;\; \tan θ=\tan 45^{∘}

θ=45^{∘}

Now, value of

\sec ^{2} θ+cosec^{2} θ

\;\; =\sec ^{2} 45^{∘}+cosec^{2} 45^{∘}

\;\; =(√{2})^{2}+(√{2})^{2}

\;\; =2+2=4

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