If $\sin \theta=\frac{3}{4}$, show that $\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3}$.
$\mathrm{LHS}=\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}$
$=\sqrt{\frac{1}{\tan ^{2} \theta}}$
$=\sqrt{\cot ^{2} \theta}$
$=\cot \theta$
$=\sqrt{\operatorname{cosec}^{2} \theta-1}$
$=\sqrt{\left(\frac{1}{\sin \theta}\right)^{2}-1}$
$=\sqrt{\left(\frac{1}{\left(\frac{3}{4}\right)}\right)^{2}-1}$
$=\sqrt{\left(\frac{4}{3}\right)^{2}-1}$
$=\sqrt{\frac{16}{9}-1}$
$=\sqrt{\frac{16-9}{9}}$
$=\sqrt{\frac{7}{9}}$
$=\frac{\sqrt{7}}{3}$
$=\mathrm{RHS}$
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