Please submit your concern

Question:

If 0° < θ < 90°, $\sqrt{\frac{se{c}^2\theta +\cos e{c}^2\theta }{{\tan }^2\theta -{\sin }^2\theta }}$ is equal to:

Explanation:

= $\sqrt{\frac{se{c}^2\theta +\cos e{c}^2\theta }{{\tan }^2\theta -{\sin }^2\theta }}$ = $\sqrt{\frac{{\displaystyle \frac{1}{{\cos }^2\theta }}+{\displaystyle \frac{1}{{\sin }^2\theta }}}{{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}-{\sin }^2\theta }}$

= $\sqrt{\frac{{\displaystyle \frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta {\sin }^2\theta }}}{{\displaystyle \frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\cos }^2\theta }}}}$

= $\sqrt{\frac{1}{{\cos }^2\theta {\sin }^2\theta }\times \frac{{\cos }^2\theta }{{\sin }^2\theta (1-{\cos }^2\theta )}}$

= $\sqrt{\frac{1}{{\sin }^2\theta }\times \frac{1}{{\sin }^2\theta \left({\sin }^2\theta \right)}}$

= $\sqrt{\frac{1}{{\sin }^6\theta }}$

= $\sqrt{\cos e{c}^6\theta }$

= cosec³ θ

Hence, the correct answer is Option B