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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