Question

If $\tan \left(\pi \cos \theta \right)=\cot \left(\pi \sin \theta \right)$, $0<\theta <\frac{3\pi }{4}$, then $\sin (\theta +\frac{\pi }{4})=$

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $-\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}$

$\to$ let $\pi cos\theta =A,\pi sin\theta =B$

$\Rightarrow tanA=cotB\Rightarrow \frac{sinA}{cosA}=\frac{cosB}{sinA}\Rightarrow cosAcosB-sinAsinB=0$

$\Rightarrow cos(A+B)=0\Rightarrow A+B=\pi /2=\pi (sin\theta +cos\theta )=\frac{\pi }{2}$

$\Rightarrow sin\theta +cos\theta =\frac{1}{2}$

Now $sin(\theta +\pi /4)=sin\theta cos\pi /4+cos\theta sin\theta \pi /4=\frac{sin\theta }{\sqrt{2}}+\frac{cos\theta }{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}\left(\frac{1}{2}\right)=\frac{1}{2\sqrt{2}}\Rightarrow \left(C\right)$