Question
If $\cos\theta=\frac{1}{\sqrt{3}},$ show that $\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{3}{5}.$
✓
Answer
$\cot\theta=\frac{1}{\sqrt{3}},$
Prove that $\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{3}{5}$
$\cot\theta=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{1}{\sqrt{3}}$
Let x be the hypotenuse
By appling Pythagoras
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{x}^2=(\sqrt{3})^2+1$
$\text{x}^2=3+1$
$\text{x}^2=3+1 \Rightarrow\ \text{x}=2$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{1}{2}$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$
$\frac{1-\cos^2\theta}{2-\sin^2\theta}\Rightarrow\frac{1-\Big(\frac{1}{2}\Big)^2}{2-\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$\Rightarrow\frac{1-\frac{1}{4}}{2-\frac{3}{4}}\Rightarrow\frac{\frac{3}{4}}{\frac{5}{4}}$
$=\frac{3}{5}$
Prove that $\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{3}{5}$
$\cot\theta=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{1}{\sqrt{3}}$
Let x be the hypotenuse
By appling Pythagoras
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{x}^2=(\sqrt{3})^2+1$
$\text{x}^2=3+1$
$\text{x}^2=3+1 \Rightarrow\ \text{x}=2$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{1}{2}$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$
$\frac{1-\cos^2\theta}{2-\sin^2\theta}\Rightarrow\frac{1-\Big(\frac{1}{2}\Big)^2}{2-\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$\Rightarrow\frac{1-\frac{1}{4}}{2-\frac{3}{4}}\Rightarrow\frac{\frac{3}{4}}{\frac{5}{4}}$
$=\frac{3}{5}$
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