If \(\sin\theta-\cos\theta=0\), then the value of \(\sin^4\theta+\cos^4\theta\) is
1
\(\dfrac{3}{4}\)
\(\dfrac{1}{2}\)
\(\dfrac{1}{4}\)

Step 1: We are given that \(\sin\theta - \cos\theta = 0\).
This means \(\sin\theta = \cos\theta\).
Step 2: From trigonometry, we know that:
\(\sin^2\theta + \cos^2\theta = 1\) (Pythagoras identity).
Step 3: Since \(\sin\theta = \cos\theta\), let each of them be equal to \(x\).
So, \(\sin\theta = x\) and \(\cos\theta = x\).
Step 4: Substitute into the identity:
\(x^2 + x^2 = 1\)
\(2x^2 = 1\)
\(x^2 = \dfrac{1}{2}\)
\(x = \dfrac{1}{\sqrt{2}}\)
Step 5: Now calculate the required expression:
\(\sin^4\theta + \cos^4\theta = x^4 + x^4 = 2x^4\)
Step 6: Since \(x = \dfrac{1}{\sqrt{2}}\):
\(x^4 = \left(\dfrac{1}{\sqrt{2}}\right)^4 = \dfrac{1}{4}\)
So, \(2x^4 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}\).
Final Answer: \(\sin^4\theta + \cos^4\theta = \dfrac{1}{2}\).