Question
If $\sin\theta=\frac{3}{4},$ show that $\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}=\frac{\sqrt{7}}{3}.$
✓
Answer
$\sin\theta=\frac34\Rightarrow\text{cosec}\theta=\frac43$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac34$
Let $BC = 3$ and $AC = 4$
Then, by pythagoras theorem,
$AB^2 = AC^2 - BC^2$
$\Rightarrow 4^2 - 3^2 = 16 - 9 = 7$
$\Rightarrow\text{AB}=\sqrt{7}$
Now,
$\cot\theta=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{\sqrt{7}}{3}$
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{4}{\sqrt{7}}$
$\therefore\text{L.H.S.}=\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}$
$=\sqrt{\frac{\big(\frac{4}{3}\big)^2-\Big(\frac{\sqrt{7}}{3}\Big)}{\Big(\frac{4}{\sqrt{7}}\Big)^2-1}}$
$=\sqrt{\frac{\frac{\frac{16}{9}-\frac{7}{9}}{12}}{\frac{16}{7}-1}}$
$=\sqrt{\frac{1}{\frac97}}$
$=\sqrt{\frac{7}{9}}$
$=\frac{\sqrt{7}}{3}$
$=\text{R.H.S.}$
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