SSC CGL 23 Aug 2021 Shift 2 Paper

Question : 72 of 100

Marks: +1, -0

0^{∘} < θ < 90^{∘}, √{\;{\sec ^{2}+cosec^{2} θ}/{\tan ^{2} θ-\sin ^{2} θ}}

Solution:

√{\;{\sec ^{2} θ+cosec^{2} θ}/{\tan ^{2} θ-\sin ^{2} θ}}

=√{\;{\;{1}/{\cos ^{2} θ}+{1}/{\sin ^{2} θ}}/{{\sin ^{2} θ}/{\cos ^{2} θ}-\sin ^{2} θ}}

=√{\;{\;{\sin ^{2} θ+\cos ^{2} θ}/{\sin ^{2} θ \cos ^{2} θ}}/{{\sin ^{2} θ-\sin ^{2} θ \cos ^{2} θ}/{\cos ^{2} θ}}}

=√{\;{\;{1}/{\sin ^{2} θ}}/{{\sin ^{2} θ(1-\cos ^{2} θ)}/{1}}}

\;\; =√{\;{1}/{\sin ^{2} θ × \sin ^{2} θ × \sin ^{2} θ}}

\;\; =√{cosec^{6} θ}

\;\; =cosec^{3} θ

The value of given expression is

cosec^{3} θ

Go to Question:

Prev Question

Next Question