MCQ
$\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}= ?$
- A$\tan \theta+\cot \theta$
- ✓$\tan \theta-\cot \theta$
- C$\sec \theta+\operatorname{cosec} \theta$
- DNone of these
✓
Answer
Correct option: B.
$\tan \theta-\cot \theta$
We have, $\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}$
$=\sqrt{\left(1+\tan ^2 \theta\right)+\left(1+\cot ^2 \theta\right)}$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta+2}$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta+2 \tan \theta \cdot \cot \theta}\ [\because \tan \theta \cdot \cot \theta=1]$
$=\sqrt{(\tan \theta+\cot \theta)^2}$
$=(\tan \theta+\cot \theta)$
$=\sqrt{\left(1+\tan ^2 \theta\right)+\left(1+\cot ^2 \theta\right)}$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta+2}$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta+2 \tan \theta \cdot \cot \theta}\ [\because \tan \theta \cdot \cot \theta=1]$
$=\sqrt{(\tan \theta+\cot \theta)^2}$
$=(\tan \theta+\cot \theta)$
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