Question
If $\cot\theta=\frac{1}{\sqrt{3}},$ write the value of $\frac{1-\cos^2\theta}{2-\sin^2\theta}.$
✓
Answer
We have, $\cot\theta=\frac{1}{\sqrt{3}}$ In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(\sqrt{3})^2+(1)^2$ $\Rightarrow\text{AC}^2=3+1$ $\Rightarrow\text{AC}^2=4$ $\Rightarrow\text{AC}^2=4$ $\therefore\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{1}{2}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$ Now, $\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{1-\Big(\frac{1}{2}\Big)^2}{2-\Big(\frac{\sqrt{3}}{2}\Big)^2}$ $=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$ $=\frac{3}{5}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(\sqrt{3})^2+(1)^2$ $\Rightarrow\text{AC}^2=3+1$ $\Rightarrow\text{AC}^2=4$ $\Rightarrow\text{AC}^2=4$ $\therefore\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{1}{2}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$ Now, $\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{1-\Big(\frac{1}{2}\Big)^2}{2-\Big(\frac{\sqrt{3}}{2}\Big)^2}$ $=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$ $=\frac{3}{5}$
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