Question

If $\sin \theta =\frac{3}{4}$, show that $\sqrt{\frac{{\mathrm{cosec}}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{7}}{3}$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{{\mathrm{cosec}}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}} \\ =\sqrt{\frac{1}{{\tan }^2\theta }} \\ =\sqrt{{\cot }^2\theta } \\ =\cot \theta \\ =\sqrt{{\mathrm{cosec}}^2\theta -1} \\ =\sqrt{{\left(\frac{1}{\sin \theta }\right)}^2-1} \\ =\sqrt{{\left(\frac{1}{\left({\displaystyle \frac{3}{4}}\right)}\right)}^2-1} \\ =\sqrt{{\left(\frac{4}{3}\right)}^2-1} \\ =\sqrt{\frac{16}{9}-1} \\ =\sqrt{\frac{16-9}{9}} \\ =\sqrt{\frac{7}{9}} \\ =\frac{\sqrt{7}}{3} \\ =\mathrm{RHS} \end{gathered}$$