MCQ
If $\tan\theta=\frac{1}{\sqrt7}$ then $\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=? $
- A$\frac{-2}{3}$
- B$\frac{-3}{4}$
- C$\frac{2}{3}$
- ✓$\frac{3}{4}$
✓
Answer
Correct option: D.
$\frac{3}{4}$
$\tan\theta=\frac{1}{\sqrt7}$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\sqrt7$
Now, $\sec^2\theta=(1+\tan^2\theta)$
$1+\Big(\frac{1}{\sqrt7}\Big)^2=1+\frac{1}{7}=\frac{8}{7}$
And, $\text{cosec}^2\theta=(1+\cot^2\theta)$
$=1+\big(\sqrt7\big)^2=1+7=8$
$\therefore\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}=\frac{48}{64}=\frac{3}{4}$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\sqrt7$
Now, $\sec^2\theta=(1+\tan^2\theta)$
$1+\Big(\frac{1}{\sqrt7}\Big)^2=1+\frac{1}{7}=\frac{8}{7}$
And, $\text{cosec}^2\theta=(1+\cot^2\theta)$
$=1+\big(\sqrt7\big)^2=1+7=8$
$\therefore\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}=\frac{48}{64}=\frac{3}{4}$
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