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
Given,$\cos\theta=\frac{3}{4}$
base = 3, perpendicular = 4
by P.G.T.
$\text{h}=\sqrt{\text{P}^2+\text{b}^2}=\sqrt{4^2+3^2}$
$=\sqrt{16+9}=\sqrt{25}=5$
$\sec\theta=\frac{\text{h}}{\text{b}}=\frac{5}{3},\text{cosec}\theta=\frac{\text{h}}{\text{P}}=\frac{5}{4}$
Taking L.H.S. $\sqrt{\frac{\sec\theta-\text{cosec}\theta}{\sec+\text{cosec}\theta}}=\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}}$
$=\sqrt{\frac{\frac{20-15}{12}}{\frac{20+15}{12}}}=\sqrt{\frac{5}{35}}=\sqrt{\frac{1}{7}}\ =\text{R.H.S}$
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