MCQ
$\sec \theta$ when expresed in terms of $\cot \theta$, is equal to
- A$\frac{1+\cot ^2 \theta}{\cot \theta}$
- B$\sqrt{1+\cot ^2 \theta}$
- ✓$\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
- D$\frac{\sqrt{1-\cot ^2 \theta}}{\cot \theta}$
✓
Answer
Correct option: C.
$\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
(C)$\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}$
$
\begin{array}{l}
\text { } 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \Rightarrow \sqrt{1+\cot ^2 \theta}=\operatorname{cosec} \theta \\
\qquad \frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}=\frac{\operatorname{cosec} \theta}{\cot \theta}=\frac{1}{\sin \theta} \times \frac{\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}=\sec \theta
\end{array}
$
$
\begin{array}{l}
\text { } 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \Rightarrow \sqrt{1+\cot ^2 \theta}=\operatorname{cosec} \theta \\
\qquad \frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}=\frac{\operatorname{cosec} \theta}{\cot \theta}=\frac{1}{\sin \theta} \times \frac{\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}=\sec \theta
\end{array}
$
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