Question

Solve $tan(\frac{\pi }{2}cos\theta )=cot(\frac{\pi }{2}sin\theta )$

• A. $\theta =(2k+3)\frac{\pi }{2}$, where $kϵz$
• B. $\theta =k\pi$, where $kϵz$
• C. $\theta =2k\pi$, where $kϵz$
• D. $\theta =(2k+1)\frac{\pi }{2}$, where $kϵz$

Answer 1

Answer: (D)

$\theta =(2k+1)\frac{\pi }{2}$, where $kϵz$

$\tan \left(\frac{\pi }{2}\cos \theta \right)=\cot \left(\frac{\pi }{2}\sin \theta \right)=\tan (\frac{\pi }{2}-\frac{\pi }{2}\tan \theta )$

$\Rightarrow \frac{\pi }{2}\cos \theta =\frac{\pi }{2}-\frac{\pi }{2}\sin \theta \Rightarrow \frac{\pi }{2}\left(\cos \theta +\sin \theta \right)=\frac{\pi }{2}$

$\Rightarrow \sqrt{2}\sin (\theta +\frac{\pi }{4})=1\Rightarrow \sin (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

$\Rightarrow \theta +\frac{\pi }{4}=n\pi +{(-1)}^n\left(\frac{\pi }{4}\right)\Rightarrow \theta =(2k+1)\frac{\pi }{2}$ where $k\in I$