SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 70 of 100

Marks: +1, -0

If

\;{\sin θ+\cos θ}/{\sin θ-\cos θ}=3

, then the value of

\sin ^{4} θ-\cos ^{4} θ

is equal to

$\;{1}/{5}$
$\;{2}/{5}$
$\;{3}/{5}$
$\;{4}/{5}$

Solution:

Given,

\;{\sin θ+\cos θ}/{\sin θ-\cos θ}=3

⇒ \;\; \sin θ+\cos θ=3 \sin θ-3 \cos θ

⇒\;3 \sin θ-\sin θ=\cos θ+3 \cos θ

⇒\;2 \sin θ=4 \cos θ

⇒\; \;\; \;{\sin θ}/{\cos θ}=\;{4}/{2}=2

⇒ \;\; \tan θ=2=\;{\;\text " Perpendicular "\;}/{\;\text " Base "\;}

∴ \;\;

Hypotenuse

=√{(2)^{2}+(1)^{2}}=√{5}

∴ \;\; \sin θ=\;{\;\text " Perpendicular "\;}/{\;\text " Hypotenuse "\;}=\;{2}/{√{5}}

and

\cos θ=\;{\;\text " Base "\;}/{\;\text " Hypotenuse "\;}=\;{1}/{√{5}}

Now, the value of

\sin ^{4} θ-\cos ^{4} θ

\;\; =(\;{2}/{√{5}})^{4}-(\;{1}/{√{5}})^{4}

\;\; =\;{16}/{25}-\;{1}/{25}=\;{16-1}/{25}=\;{15}/{25}=\;{3}/{5}

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