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SSC CGL 23 Aug 2021 Shift 1 Paper

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Question : 58 of 100

Marks: +1, -0

If

\sin^{2}\theta-\cos^{2}\theta-3\sin\theta+2

0^{\circ} < \theta<90^{\circ}

, then what is the value of

\frac{1}{\sqrt{\sec\theta-\tan\theta}}

is

$\sqrt[2]{3}$
$\sqrt[2]{2}$
$\sqrt[4]{3}$
$\sqrt[4]{2}$

Solution:

\sin^{2}\theta-\cos^{2}\theta-3\sin\theta+2=0

\sin^{2}\theta-(1-\sin^{2}\theta)-3\sin\theta+2=0

[\because \sin^{2}\theta+\cos^{2}\theta=1]

\Rightarrow \sin^{2}\theta-1+\sin^{2}\theta-3\sin\theta+2=0

\Rightarrow 2\sin^{2}\theta-3\sin\theta+1=0

\Rightarrow 2\sin^{2}\theta-2\sin\theta-\sin\theta+1=0

\Rightarrow 2\sin\theta(\sin\theta-1)-1(\sin\theta-1)=0

\Rightarrow (2\sin\theta-1)(\sin\theta-1)=0

\Rightarrow \sin\theta=\frac{1}{2},\ \sin\theta=1

and

\sin\theta=\sin30^{\circ}

\sin\theta=\sin90^{\circ}

\theta=30^{\circ},\ \theta=90^{\circ}

\theta=90^{\circ}

will not take because value of

\theta

varies between

0^{\circ}

and

90^{\circ}

.

Then,

\frac{1}{\sqrt{\sec\theta-\tan\theta}}

=\frac{1}{\sqrt{\sec30^{\circ}-\tan30^{\circ}}}

=\frac{1}{\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right)^{1/2}}

=\frac{1}{\left[\frac{1}{\sqrt{3}}\right]^{1/2}}=(3)^{1/4}\ \text{or}\ \sqrt[4]{3}

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