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} $

Given:

\( \sin \theta-\cos \theta=0 \)

To do:

We have to find the value of \( \left(\sin ^{4} \theta+\cos ^{4} \theta\right) \).

Solution:

We know that,

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

$\tan 45^{\circ}=1$

Therefore,

$\sin \theta-\cos \theta=0$

$\sin \theta=\cos \theta$

$\frac{\sin \theta}{\cos \theta}=1$

$\tan \theta=1$

$\tan \theta=\tan 45^{\circ}$

$\Rightarrow \theta=45^{\circ}$

This implies,

$\sin ^{4} \theta+\cos ^{4} \theta=\sin ^{4} 45^{\circ}+\cos ^{4} 45^{\circ}$

$=(\frac{1}{\sqrt{2}})^{4}+(\frac{1}{\sqrt{2}})^{4}$         (Since $\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$)

$=\frac{1}{4}+\frac{1}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$