Question
If $\tan\theta=\frac{1}{\sqrt{7}},$ show that $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}=\frac{3}{4}.$
✓
Answer
We have,
$\tan\theta=\frac{1}{\sqrt{7}}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(1)^2+(\sqrt{7})^2$
$\Rightarrow\text{AC}^2=8$
$\Rightarrow\text{AC}=2\sqrt{2}$
$\therefore\text{cosec }\theta=\frac{2\sqrt{2}}{1},$ and $\sec\theta=\frac{2\sqrt{2}}{\sqrt{7}}$
Now, $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}=\frac{(2\sqrt{2})^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{(2\sqrt{2})^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}$
$=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\big(\frac{56-8}{7}\big)}{\big(\frac{56+8}{7}\big)}$
$=\frac{48}{64}$
$=\frac{6}{8}=\frac{3}{4}$
$\tan\theta=\frac{1}{\sqrt{7}}$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(1)^2+(\sqrt{7})^2$
$\Rightarrow\text{AC}^2=8$
$\Rightarrow\text{AC}=2\sqrt{2}$
$\therefore\text{cosec }\theta=\frac{2\sqrt{2}}{1},$ and $\sec\theta=\frac{2\sqrt{2}}{\sqrt{7}}$
Now, $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}=\frac{(2\sqrt{2})^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{(2\sqrt{2})^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}$
$=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\big(\frac{56-8}{7}\big)}{\big(\frac{56+8}{7}\big)}$
$=\frac{48}{64}$
$=\frac{6}{8}=\frac{3}{4}$
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