Compound And Multiple Angles

Question

If $\cos	heta = \frac{3}{5}$ with $	heta$ in the first quadrant, then applying the half-angle identity gives the value of $	an\frac{	heta}{2}$ equal to which of the following?

RankGuru · Accepted Answer

The controlling tool is the half-angle identity $	an\frac{	heta}{2} = \frac{1 - \cos	heta}{\sin	heta}$, 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 $	heta$ lies in the first quadrant, both its sine and cosine are positive, so from $\cos	heta = \frac{3}{5}$ we get $\sin	heta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$. Substituting into the identity gives $	an\frac{	heta}{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 $\frac{\sin	heta}{1 + \cos	heta} = \frac{4/5}{8/5} = \frac{4}{8} = \frac{1}{2}$ 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 $	an	heta = \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 $	an	heta$ for $\cos	heta = \frac{3}{5}$, so the half-angle value is verified.

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Compound And Multiple Angles

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