Question
If $\tan\theta=\frac{1}{\sqrt{7}},$ show that $\frac{\big(\text{coses}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\frac34.$
✓
Answer
Given: $\tan\theta=\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{7}}$
Let BC = 1k and $\text{AB}=\sqrt{7}\text{k}$
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB}^2 + \text{BC}^2)$
$\Rightarrow\text{AC}^2=\Big[\big(\sqrt{7}\text{k}\big)^2+\big(1\text{k}\big)^2\Big]$
$= 7\text{k}^2 + 1\text{k}^2 = 8\text{k}^2$
$\Rightarrow\text{AC}=2\sqrt{2}\text{k}$
$\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{2\sqrt{2}\text{k}}{1\text{k}}={2\sqrt{2}}$
$\sec\theta=\frac{\text{AC}}{\text{AB}}=\frac{2\sqrt{2}\text{k}}{\sqrt{7}\text{k}}=\frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\begin{bmatrix}\frac{\big(2\sqrt{2}\big)^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{\big(2\sqrt{2}\big)^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}\end{bmatrix}$
$=\frac{\big(8-\frac{8}{7}\big)}{\big(8+\frac87\big)}=\frac{\big(\frac{48}{7}\big)}{\big(\frac{64}{7}\big)}=\frac{48}{64}=\frac34$
Hence, $\Big(\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}\Big)=\frac34$
Let BC = 1k and $\text{AB}=\sqrt{7}\text{k}$
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB}^2 + \text{BC}^2)$
$\Rightarrow\text{AC}^2=\Big[\big(\sqrt{7}\text{k}\big)^2+\big(1\text{k}\big)^2\Big]$
$= 7\text{k}^2 + 1\text{k}^2 = 8\text{k}^2$
$\Rightarrow\text{AC}=2\sqrt{2}\text{k}$
$\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{2\sqrt{2}\text{k}}{1\text{k}}={2\sqrt{2}}$
$\sec\theta=\frac{\text{AC}}{\text{AB}}=\frac{2\sqrt{2}\text{k}}{\sqrt{7}\text{k}}=\frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\begin{bmatrix}\frac{\big(2\sqrt{2}\big)^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{\big(2\sqrt{2}\big)^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}\end{bmatrix}$
$=\frac{\big(8-\frac{8}{7}\big)}{\big(8+\frac87\big)}=\frac{\big(\frac{48}{7}\big)}{\big(\frac{64}{7}\big)}=\frac{48}{64}=\frac34$
Hence, $\Big(\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}\Big)=\frac34$
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