Question

If $\theta \in \left(\frac{\pi }{4},\frac{\pi }{2}\right)$ and $f\left(\theta \right)=\sec 2\theta -\tan 2\theta$, then $f(\frac{\pi }{4}-\theta )$ equals

• A. $\tan \theta$
• B. $\cot \theta$
• C. $\sec 2\theta$
• D. $\tan 2\theta$

Answer 1

The correct option is A $\tan \theta$

$f\left(\theta \right)=\sec 2\theta -\tan 2\theta$

$=\frac{1-\sin 2\theta }{\cos 2\theta }$

$=\frac{{(\cos \theta -\sin \theta )}^2}{{\cos }^2\theta -{\sin }^2\theta }$

$=\frac{(\cos \theta -\sin \theta )}{\cos \theta +\sin \theta }$

$=\frac{1-\tan \theta }{1+\tan \theta }=\tan (\frac{\pi }{4}-\theta )$

$\Rightarrow f\left(\theta \right)$$=\tan (\frac{\pi }{4}-\theta )$

$\therefore f(\frac{\pi }{4}-\theta )=\tan (\frac{\pi }{4}-\frac{\pi }{4}+\theta )=\tan \theta$