SSC CHSL 12 Oct 2020 Shift 2 Paper

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Question : 65 of 100

Marks: +1, -0

If

4 \cos ^{2} \θ-3 \sin ^{2} \θ+2=0

, then the value of

\tan \θ

is (where

0 \≤ \θ < 90°

)

1
$\√{6}$
$\√{2}$
${1}/{\√{3}}$

Solution:

4 \cos ^{2} \θ-3 \sin ^{2} \θ+2=0

⇒

4(1-\sin ^{2} \θ)-3 \sin ^{2} \θ+2=0

⇒

4-4 \sin ^{2} \θ-3 \sin ^{2} \θ+2=0

⇒

-7 \sin ^{2} \θ=-6

⇒

\sin ^{2} \θ={6}/{7}

{\cosec}^{2} \θ={7}/{6}

Now, we know that
⇒

{\cosec}^{2} \θ-1=\cot ^{2} \θ

⇒

\cot ^{2} \θ={7}/{6}-1

⇒

\cot ^{2} \θ={1}/{6}

⇒

\cot \θ={1}/{\√{6}}

⇒

\tan \θ=\√{6}

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