Question
If $\cot\theta=\frac{3}{4},$ prove that $\sqrt{\frac{\sec\theta-\text{cosec }\theta}{\sec\theta+\text{cosec }\theta}}=\frac{1}{\sqrt{7}}.$
✓
Answer
$\cot\theta=\frac{3}{4}\ \text{P.T}\ \sqrt{\frac{\sec\theta-\text{cosec }\theta}{\sec\theta+\text{cosec }\theta}}=\frac{1}{\sqrt{7}}$
$\cot\theta=\frac{\text{adjacent side }}{\text{opposite side}}$
Let x be the hypotenuse by applying Pythagoras theorem.
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{x}^2=16+9$
$\text{x}^2=25\Rightarrow\text{x}=5$
$\sec\theta=\frac{\text{AC}}{\text{BC}}=\frac{5}{3}$
$\text{cosec }\theta=\frac{\text{AC}}{\text{AB}}=\frac{5}{4}$
On substituting in equation we get
$ \sqrt{\frac{\sec\theta-\text{cosec }\theta}{\sec\theta+\text{cosec }\theta}}=\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}}$
$=\sqrt{\frac{\frac{20-15}{12}}{\frac{20+15}{12}}}=\sqrt{\frac{5}{35}}=\frac{1}{\sqrt{7}}$
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