(\sqrt{\sec ^{2} \theta+cosec^{2} \theta}) (\frac{\sin \theta(1+\cos \theta)}{1+\cos \theta-\sin ^{2} \theta}), 0^{\circ}

SSC CGL 17 Aug 2021 Shift 1 Paper

Question : 65 of 100

Marks: +1, -0

(√{\sec ^{2} θ+cosec^{2} θ})

(\;{\sin θ(1+\cos θ)}/{1+\cos θ-\sin ^{2} θ}), 0^{∘} < θ < 90^{∘}

Solution:

√{\sec ^{2} θ+cosec^{2} θ}

×(\;{\sin θ(1+\cos θ)}/{1+\cos θ-\sin ^{2} θ})

[ ∵ 0^{∘} < θ < 90^{∘}]

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

×(\;{\sin θ+\sin θ \cos θ}/{\cos ^{2} θ+\cos θ})

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

×(\;{\sin θ(1+\cos θ)}/{\cos θ(1+\cos θ)})

=\;{1}/{\cos θ \sin θ} × \;{\sin θ}/{\cos θ}

=\;{1}/{\cos ^{2} θ}=\sec ^{2} θ \;\text ". "\;

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