Question:
If \(\tan \theta + \cot \theta = 2\), find the value of \(\sin^4 \theta + \cos^4 \theta\).
यदि \(\tan \theta + \cot \theta = 2\) है, तो \(\sin^4 \theta + \cos^4 \theta\) का मान ज्ञात कीजिए।
Options:
- \(\frac{1}{2}\)
- \(1\)
- \(\frac{3}{4}\)
- \(\frac{1}{4}\)
- \(\frac{1}{2}\)
- \(1\)
- \(\frac{3}{4}\)
- \(\frac{1}{4}\)
Correct Answer: a)
Explanation:
Given \(\tan \theta + \frac{1}{\tan \theta} = 2 \implies (\tan \theta – 1)^2 = 0 \implies \tan \theta = 1\).
Thus, \(\theta = 45^\circ\).
\(\sin 45^\circ = \frac{1}{\sqrt{2}}\), \(\cos 45^\circ = \frac{1}{\sqrt{2}}\).
\(\sin^4 45^\circ + \cos^4 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\).
Thus, \(\theta = 45^\circ\).
\(\sin 45^\circ = \frac{1}{\sqrt{2}}\), \(\cos 45^\circ = \frac{1}{\sqrt{2}}\).
\(\sin^4 45^\circ + \cos^4 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\).
\(\tan \theta + \cot \theta = 2 \implies \theta = 45^\circ\)
\(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
\(\sin^4 45^\circ + \cos^4 45^\circ = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)।
\(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
\(\sin^4 45^\circ + \cos^4 45^\circ = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)।