RRB Group D 17-Sep-2018 Shift 1 Solved Paper

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Question : 4 of 100

Marks: +1, -0

If

\sin ϕ-\cos ϕ =0,

then the value

of \sin ^{4} ϕ +\cos ^{4} ϕ

+\tan ^{2} ϕ

is :

$3 / 2$
2
$5 / 4$
$7 / 4$

Solution:

since

\sin ϕ-\cos ϕ =0 ⇒ \sin ϕ =\cos ϕ

⇒ ϕ =45^{∘}

Now

\sin ^{4} ϕ +\cos ^{4} ϕ +\tan ^{2} ϕ

=(\;{1}/{√{2}})^{4}+(\;{1}/{√{2}})^{4}+(1)^{2}

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

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