Search any question & find its solution
Question:
Answered & Verified by Expert
$\sec 2 \theta-\tan 2 \theta=$
Options:
Solution:
1824 Upvotes
Verified Answer
The correct answer is:
$\tan \left(\frac{\pi}{4}-\theta\right)$
$\begin{aligned} \sec 2 \theta-\tan 2 \theta &=\frac{1}{\cos 2 \theta}-\frac{\sin 2 \theta}{\cos 2 \theta}=\frac{1-\sin 2 \theta}{\cos 2 \theta} \\ &=\frac{(\cos \theta-\sin \theta)^{2}}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{(\cos \theta-\sin \theta)^{2}}{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)} \end{aligned}$
$=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}$
$=\tan \left(\frac{\pi}{4}-\theta\right)$
$=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}$
$=\tan \left(\frac{\pi}{4}-\theta\right)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.