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SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 70 of 100

Marks: +1, -0

If

\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3

, then the value of

\sin^4\theta-\cos^4\theta

is equal to

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{4}{5}$

Solution:

Given,

\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3

\Rightarrow \sin\theta+\cos\theta=3\sin\theta-3\cos\theta

\Rightarrow 3\sin\theta-\sin\theta=\cos\theta+3\cos\theta

\Rightarrow 2\sin\theta=4\cos\theta

\Rightarrow \frac{\sin\theta}{\cos\theta}=\frac{4}{2}=2

\Rightarrow \tan\theta=2=\frac{\text{ Perpendicular }}{\text{ Base }}

\therefore

Hypotenuse

=\sqrt{(2)^2+(1)^2}=\sqrt{5}

\therefore \sin\theta = \frac{\text{ Perpendicular }}{\text{ Hypotenuse }} = \frac{2}{\sqrt{5}}

and

\cos\theta = \frac{\text{ Base }}{\text{ Hypotenuse }} = \frac{1}{\sqrt{5}}

Now, the value of

\sin^4\theta-\cos^4\theta

= \left(\frac{2}{\sqrt{5}}\right)^4 - \left(\frac{1}{\sqrt{5}}\right)^4

= \frac{16}{25} - \frac{1}{25} = \frac{16-1}{25} = \frac{15}{25} = \frac{3}{5}

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