Question

If $\sin \left(\frac{\pi }{4}\cot \theta \right)=\cos \left(\frac{\pi }{4}\tan \theta \right),then\theta$=

• A. $n\pi \pm \frac{\pi }{4}$
• B. $2n\pi \pm \frac{\pi }{4}$
• C. $n\pi -\frac{\pi }{4}$
• D. $2n\pi \pm \frac{\pi }{6}$

Answer 1

Answer: (A)

$\sin \left(\frac{\pi }{4}\cot \theta \right)=\cos \left(\frac{\pi }{4}\tan \theta \right)$
$\Rightarrow \sin \left(\frac{\pi }{4}\tan \theta \right)=\cos \left(\frac{\pi }{2}-\frac{\pi }{4}\cot \theta \right)$
$\Rightarrow \frac{\pi }{4}\tan \theta =\frac{\pi }{2}-\frac{\pi }{4}\cot \theta$
$\Rightarrow \frac{\pi }{4}(\tan \theta +\cot \theta )=\frac{\pi }{2}$
$\Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=2$
$\Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =0$
$\Rightarrow (\sin \theta -\cos \theta {)}^2=0\Rightarrow \sin \theta -\cos \theta =0$
$\Rightarrow \tan \theta =1=\frac{\pi }{4}\Rightarrow \theta =n\pi \pm \frac{\pi }{4}$ where $n\in I$