Practicing Success
If $0^{\circ}<\theta<90^{\circ}, \sqrt{\frac{\sec ^2 \theta+{cosec}^2 \theta}{\tan ^2 \theta-\sin ^2 \theta}}$ is equal to: |
$\sec ^3 \theta$ ${cosec}^3 \theta$ $\sin ^2 \theta$ $\sec ^2 \theta$ |
${cosec}^3 \theta$ |
\(\sqrt { \frac{sec²θ + cosec²θ}{tan²θ - sin²θ}\}\) = \(\sqrt { \frac{1/cos²θ + 1/sin²θ}{sin²θ/cos²θ - sin²θ}\}\) = \(\sqrt { \frac{(cos²θ+sin²θ)/cos²θ.sin²θ }{ ( sin²θ - sin²θ.cos²θ) /cos²θ}\}\) { 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 |