If $\sin \theta-\cos \theta=0$, then the value of $\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is
(a) 1
(b) $\frac{3}{4}$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$
(c) Given, $\sin \theta-\cos \theta=0$
$\Rightarrow$ $\sin \theta=\cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=1$
$\Rightarrow$ $\tan \theta=1$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\tan 45^{\circ}=1\right]$
$\Rightarrow$ $\tan \theta=\tan 45^{\circ}$
$\therefore$ $\theta=45^{\circ}$
Now, $\sin ^{4} \theta+\cos ^{4} \theta=\sin ^{4} 45^{\circ}+\cos ^{4} 45^{\circ}$
$=\left(\frac{1}{\sqrt{2}}\right)^{4}+\left(\frac{1}{\sqrt{2}}\right)^{4}$ $\left[\because \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
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