MCQ
Choose the correct answer from the given four options. If $\sin\theta-\cos\theta=0,$ then the value of $(\sin^4\theta+\cos^4\theta)$ is :
- A$1$
- B$\frac{3}{4}$
- ✓$\frac{1}{2}$
- D$\frac{1}{4}$
✓
Answer
Correct option: C.
$\frac{1}{2}$
Given $,\sin\theta-\cos\theta=0$
$\Rightarrow\ \sin\theta=\cos\theta$
$\Rightarrow\frac{\sin\theta}{\cos\theta}=1$
$\Rightarrow\ \tan\theta=1\bigg[\because\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\tan45^\circ=1\bigg]$
$\Rightarrow\ \tan\theta=\tan45^\circ$
$\therefore\ \theta=45^\circ$
Now $, \sin^4\theta+\cos^4\theta$
$=\sin^445^\circ+\cos^445^\circ$
$=\Big(\frac{1}{\sqrt{2}}\Big)^4+\Big(\frac{1}{\sqrt{2}}\Big)^4\Big[\because\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt{2}}\Big]$
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
$\Rightarrow\ \sin\theta=\cos\theta$
$\Rightarrow\frac{\sin\theta}{\cos\theta}=1$
$\Rightarrow\ \tan\theta=1\bigg[\because\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\tan45^\circ=1\bigg]$
$\Rightarrow\ \tan\theta=\tan45^\circ$
$\therefore\ \theta=45^\circ$
Now $, \sin^4\theta+\cos^4\theta$
$=\sin^445^\circ+\cos^445^\circ$
$=\Big(\frac{1}{\sqrt{2}}\Big)^4+\Big(\frac{1}{\sqrt{2}}\Big)^4\Big[\because\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt{2}}\Big]$
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
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