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