Question
If $\cot\theta=\frac{3}{4},$ show that $\sqrt{\frac{\sec\theta-\text{cosec}\theta}{\sec\theta+\text{cosec}\theta}}=\frac{1}{\sqrt{7}}.$
✓
Answer
$\cot\theta=\frac34$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\cot\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac34$
Let $AB = 3$ and $BC = 4$
Then, by pythagoras theorem,
$AC^2 = AB^2 + BC^2$
$\Rightarrow 3^2 + 4^2$
$= 9 + 16 = 25$
$\Rightarrow AC = 5$
Now,
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac53$
$\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac54$
$\therefore\text{L.H.S.}=\sqrt{\frac{\sec\theta-\text{cosec}\theta}{\sec\theta-\text{cosec}\theta}}$
$=\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{4}+\frac{5}{4}}}$
$=\sqrt{\frac{\frac{20-15}{12}}{\frac{20+15}{12}}}$
$=\sqrt{\frac{5}{35}}$
$=\sqrt{\frac{1}{7}}$
$=\frac{1}{\sqrt{7}}$
$=\text{R.H.S.}$
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