Question : If $(\sin \theta-\cos \theta)=0$, then the value of $\sin\;(\pi-\theta)+\sin \left(\frac{\pi}{2}-\theta\right)$ is:
Option 1: $1$
Option 2: $0$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{2}$
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Correct Answer: $\sqrt{2}$
Solution : $(\sin \theta-\cos \theta)=0$ $⇒\sin \theta=\cos \theta$ $⇒\frac{\sin \theta}{\cos \theta}=1$ $⇒\tan\theta=1=\tan45°$ $\therefore \theta = 45°$ $\sin\;(\pi-\theta)+\sin \left(\frac{\pi}{2}-\theta\right)$ $=\sin\;(180°-45°)+\sin (90°-45°)$ $= \sin135°+\cos 45°$ $=\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$ [$\because \sin 135°=\sin(90°+45°)=\frac{1}{\sqrt{2}}$] $ = \sqrt{2}$ Hence, the correct answer is $ \sqrt{2}$.
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