Correct option is: C) 45°
We have, sin \(\theta\) . cos \(\theta\) = \(\frac 12\)
= \(sin^2 \theta \, cos^2 \theta = \frac 14 \) (By squaring on both sides)
= \(sin^2 \theta (1-sin^2\theta) = \frac 14\)
= \(sin^2 \theta\, sin^4 \theta = \frac 14\)
= \(4\, sin^4 \theta - 4 \, sin^2\theta + 1 = 0 \) which is a quadratic equation in \(sin^2\theta\)
= \((2\, sin^2\theta -1)^2 = 0 \) ( \(\because\) \(a^2 = 2ab + b^2 = (a+b)^2\))
= \(2\, sin^2\theta -1 = 0\)
= \(sin^2\theta = \frac 12 = (\frac 1{\sqrt2})^2 = (sin\, 45^\circ)^2\) (\(\because\) \(sin\, 45^\circ = \frac 1{\sqrt2}\))
= \(\theta\) = \(45^\circ\)
Alternative method :
We have sin \(\theta\) cos \(\theta\) = \(\frac 12\)
= 2 sin \(\theta\) . cos \(\theta\) = \(\frac 22\) = 1
= \(sin^2 \theta = sin \frac \pi 2 = sin \, 90^\circ\)
= 2 \(\theta\) = \(90^\circ\)
= \(\theta\) = \(\frac {90^\circ}{2} \) = \(45^\circ\)
