\[ \text { if } 7 \sin ^2 \theta+3 \cos ^2 \theta=4,\left(0^{\circ}<\theta<90^{\circ}\right) \text {. then value of } \theta \text { is } \]

Given the equation:
\[7 \sin^2 \theta + 3 \cos^2 \theta = 4\]

Since \(\cos^2 \theta = 1 – \sin^2 \theta\), we can substitute this into the equation:
\[7 \sin^2 \theta + 3 (1 – \sin^2 \theta) = 4\]

Expanding and simplifying:
\[7 \sin^2 \theta + 3 – 3 \sin^2 \theta = 4\]
\[4 \sin^2 \theta = 1\]
\[\sin^2 \theta = \frac{1}{4}\]
\[\sin \theta = \pm \frac{1}{2}\]

Since \(0^\circ < \theta < 90^\circ\), we know that \(\sin \theta\) is positive in this interval. Therefore, we can disregard the negative solution, leaving us with:
\[\sin \theta = \frac{1}{2}\]

The value of \(\theta\) that satisfies this equation in the given interval is:
\[\theta = 30^\circ \text{ or } \frac{\pi}{6} \text{ radians}\]

So, the value of \(\theta\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.