Compound And Multiple Angles

The controlling tool is the half-angle identity $\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }$, which expresses the half-angle tangent purely through the sine and cosine of the full angle, with the quadrant fixing the sign of the sine. Since $\theta$ lies in the first quadrant, both its sine and cosine are positive, so from $\cos \theta =\frac{3}{5}$ we get $\sin \theta =\sqrt{1-\frac{9}{25}}=\frac{4}{5}$. Substituting into the identity gives $\tan \frac{\theta }{2}=\frac{1-\frac{3}{5}}{\frac{4}{5}}=\frac{\frac{2}{5}}{\frac{4}{5}}=\frac{2}{4}=\frac{1}{2}$. The equivalent form $\frac{\sin \theta }{1+\cos \theta }=\frac{4/5}{8/5}=\frac{4}{8}=\frac{1}{2}$ produces the same value, confirming consistency between the two standard half-angle expressions. Option $\frac{1}{3}$ arises from incorrectly computing the sine. Option $\frac{2}{3}$ comes from misreading the cosine value. Option $\frac{1}{4}$ results from applying a halving one extra time. This matches the canonical half-angle pattern of JEE Advanced. As a final plausibility check, using the tangent double-angle formula $\tan \theta =\frac{2t}{1-t^2}$ with $t=\frac{1}{2}$ gives $\frac{1}{1-\frac{1}{4}}=\frac{1}{3/4}=\frac{4}{3}$, which is exactly $\tan \theta$ for $\cos \theta =\frac{3}{5}$, so the half-angle value is verified.