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# If $\sin \theta+\cos \theta=1$,then the value of $\sin 2\theta$ is equal to

$\begin{array}{1 1}(A)\;1&(B)\;\large\frac{1}{2}\\(C)\;0&(D)\;-1\end{array}$

Toolbox:
• $\sin^2\theta+\cos ^2\theta=1$
• $\sin 2\theta=2\sin \theta\cos\theta$
Given :
$\sin \theta+\cos \theta=1$
Squaring both side
$(\sin \theta+\cos \theta)^2=1$
$\sin^2\theta+\cos^2\theta+2\sin \theta\cos \theta=1$
$1+2\sin \theta\cos \theta=1$
$2\sin \theta\cos\theta=0$
$\sin 2\theta=0$
Hence (C) is the correct answer.