$\sec 2 \theta-\tan 2 \theta=$
$\sec 2 \theta-\tan 2 \theta=$
- $\tan \left(\frac{\pi}{4}-\theta\right)$
- $\tan 2 \theta$
- $\cot 2 \theta$
- $\cot \left(\frac{\pi}{4}-\theta\right)$
Solution
$\begin{aligned} \sec 2 \theta-\tan 2 \theta &=\frac{1}{\cos 2 \theta}-\frac{\sin 2 \theta}{\cos 2 \theta}=\frac{1-\sin 2 \theta}{\cos 2 \theta} \\ &=\frac{(\cos \theta-\sin \theta)^{2}}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{(\cos \theta-\sin \theta)^{2}}{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)} \end{aligned}$
$=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}$
$=\tan \left(\frac{\pi}{4}-\theta\right)$
Asked in: MHT CET 2020 (12 Oct Shift 2)
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