SSC CPO SI and ASI 2016 Paper

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Question : 150 of 200

Marks: +1, -0

If 0 ˂ θ ˂ 90°, tan θ + sin θ = m, and tan θ – sin θ = n where m ≠ n, then the value of

m^2 – n^2

is

$4 mn$
2( $m^2 + n^2$ )
2 $(\tan ^{2} \θ+\sin ^{2} \θ)$
4 $√{m n}$

Solution:

(4) Here,

m+n=2 \tan \θ

and

m-n=2 \sin \θ

, So,

m^{2}-n^{2}=(m+n)(m-n)=4 \tan \θ× \sin \θ

=4 √{\tan ^{2} \θ . \sin ^{2} \θ}

=4 √{\tan ^{2} \θ\ (1-\cos ^{2} \θ)}

=4 √{\tan ^{2} \θ-\sin ^{2} \θ}

=4 √{(\tan \θ+\sin \θ)(\tan \θ-\sin \θ)}

=4 √{m n}

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