MCQ
$\sec \theta$ when expressed in terms of $\cot \theta$, is equal to:
- A$\frac{\sqrt{1-\cot ^2 \theta}}{\cot \theta}$
- B$\frac{1+\cot ^2 \theta}{\cot \theta}$
- ✓$\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
- D$\sqrt{1+\cot ^2 \theta}$
✓
Answer
Correct option: C.
$\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
As we know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\text { and } \cot \theta=\frac{1}{\tan \theta}$
$\Rightarrow \tan \theta=\frac{1}{\cot \theta}$
$\therefore \sec ^2 \theta=1+\left(\frac{1}{\cot \theta}\right)^2$
$=1+\frac{1}{\cot ^2 \theta}$
$\Rightarrow \sec ^2 \theta=\frac{\cot ^2 \theta+1}{\cot ^2 \theta}$
$\Rightarrow \sec \theta=\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
$\sec ^2 \theta=1+\tan ^2 \theta$
$\text { and } \cot \theta=\frac{1}{\tan \theta}$
$\Rightarrow \tan \theta=\frac{1}{\cot \theta}$
$\therefore \sec ^2 \theta=1+\left(\frac{1}{\cot \theta}\right)^2$
$=1+\frac{1}{\cot ^2 \theta}$
$\Rightarrow \sec ^2 \theta=\frac{\cot ^2 \theta+1}{\cot ^2 \theta}$
$\Rightarrow \sec \theta=\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
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