Question
If $\cot\theta=\sqrt{3},$ find the value of $\frac{\text{cosec}^2\theta+\cot^2\theta}{\text{cosec}^2\theta-\sec^2\theta}$.
✓
Answer
We have, $\cot\theta=\sqrt{3}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\ \text{AC}^2=(1)^2+(\sqrt{3})^2$
$\Rightarrow\ \text{AC}^2=4$
$\Rightarrow\ \text{AC}=2$
$\therefore\ \text{cosec }\theta=\frac{\text{AC}}{\text{AB}}=2,\cot\theta=\frac{\text{BC}}{\text{AB}}=\sqrt{3},\sec\theta=\frac{\text{AC}}{\text{BC}}=\frac{2}{\sqrt{3}}$
Now, $\frac{\text{cosec}^2\theta+\cot^2\theta}{\text{cosec}^2\theta-\sec^2\theta}=\frac{(2)^3+(\sqrt{3})^2}{(2)^2-\Big(\frac{2}{\sqrt{3}}\Big)^2}$
$=\frac{4+3}{4-\frac{4}{3}}$
$=\frac{7}{\frac{12-4}{3}}$
$=\frac{21}{8}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\ \text{AC}^2=(1)^2+(\sqrt{3})^2$
$\Rightarrow\ \text{AC}^2=4$
$\Rightarrow\ \text{AC}=2$
$\therefore\ \text{cosec }\theta=\frac{\text{AC}}{\text{AB}}=2,\cot\theta=\frac{\text{BC}}{\text{AB}}=\sqrt{3},\sec\theta=\frac{\text{AC}}{\text{BC}}=\frac{2}{\sqrt{3}}$
Now, $\frac{\text{cosec}^2\theta+\cot^2\theta}{\text{cosec}^2\theta-\sec^2\theta}=\frac{(2)^3+(\sqrt{3})^2}{(2)^2-\Big(\frac{2}{\sqrt{3}}\Big)^2}$
$=\frac{4+3}{4-\frac{4}{3}}$
$=\frac{7}{\frac{12-4}{3}}$
$=\frac{21}{8}$
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