If $0^{\circ}<\theta<90^{\circ}, \sqrt{\frac{\sec ^2 \theta+{cosec}^2 \theta}{\tan ^2 \theta-\sin ^2 \theta}}$ is equal to:
Answer & explanation
Correct answer: option 2
\(\sqrt { \frac{sec²θ + cosec²θ}{tan²θ - sin²θ}\}\)
= \(\sqrt { \frac{1/cos²θ + 1/sin²θ}{sin²θ/cos²θ - sin²θ}\}\)
= \(\sqrt { \frac{(cos²θ+sin²θ)/cos²θ.sin²θ }{ ( sin²θ - sin²θ.cos²θ) /cos²θ}\}\)
{ using , cos²θ+sin²θ = 1 }
= \(\sqrt { \frac{(1)/cos²θ.sin²θ }{ ( sin²θ( 1 - cos²θ) /cos²θ}\}\)
= \(\sqrt { \frac{(1)/sin²θ }{ ( sin²θ. sin²θ /cos²θ}\}\)
= \(\sqrt { \frac{1}{ ( sin6 θ}\}\)
= \(\sqrt {cosec6 θ }\)
= cosec³θ