Question

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

Answer 1

If $\sin \theta =\frac{3}{4}$

$\sqrt{\frac{{\csc }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{m}}{3}$ Find m.

$A{C}^2=A{B}^2+B{C}^2$

$16=9+B{C}^2$

$\Rightarrow BC=\sqrt{7}$

(Refer to Image)

$\sin \theta =\frac{3}{4}=\frac{AB}{AC}$

$\csc \theta =\frac{AC}{AB}=\frac{4}{3}$

$\cot \theta =\frac{\sqrt{7}}{3}=\frac{BC}{AB}$

$\sec \theta =\frac{AC}{BC}=\frac{4}{\sqrt{7}}$

$\Rightarrow \sqrt{\frac{{\csc }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=$$\sqrt{\frac{{\left(\frac{4}{3}\right)}^2-{\left(\frac{\sqrt{7}}{3}\right)}^2}{{\left(\frac{4}{\sqrt{7}}\right)}^2-1}}$

$=\sqrt{\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}}$

$=\sqrt{\frac{\frac{9}{9}}{\frac{9}{7}}}$

$=\sqrt{\frac{7}{9}}$

$=\frac{\sqrt{7}}{3}$

$\therefore m=7$