SSC CHSL 17 Mar 2020 Shift 2 Paper

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Question : 70 of 100

Marks: +1, -0

(\sin \θ+\cos \θ)^{2}=2,0° < \θ < 90°

, then the value of

\θ

is

0
${\π}/{2}$
$\π$
${\π}/{4}$

Solution:

(\sin \θ+\cos \θ)^{2}=2

⇒

\sin ^{2} \θ+\cos ^{2} \θ+2 \sin \θ\cos \θ=2

[\∵(a+b)^{2}=a^{2}+b^{2}+2 a b]

⇒

1+2 \sin \θ\cos \θ=2

(\∵\sin ^{2} \θ+\cos ^{2} \θ=1)

⇒

\sin 2 \θ=2-1

(\∵\sin 2 \θ=2 \sin \θ\cos \θ)

⇒

\sin 2 \θ=1

(\∵\sin {\π}/{2}=1)

⇒

2 \θ={\π}/{2}

⇒

\θ={\π}/{4}

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