If a line makes angles α,β and γ with positive x-axis, y-axis and z-a…

The correct answer is Option (2) → $\frac{1}{2}$

Given:

$\alpha, \beta, \gamma$ are the direction angles of a line.

Direction cosines: $l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$ satisfy:

$l^2 + m^2 + n^2 = 1$

Expression to evaluate:

$\sin^2\frac{\alpha}{2} \cdot \cos^2\frac{\alpha}{2} \;+\; \sin^2\frac{\beta}{2} \cdot \cos^2\frac{\beta}{2} \;+\; \sin^2\frac{\gamma}{2} \cdot \cos^2\frac{\gamma}{2}$

Use identity: $\sin^2\theta \cdot \cos^2\theta = \frac{\sin^2 2\theta}{4}$.

So, each term becomes:

$\sin^2\frac{\alpha}{2} \cdot \cos^2\frac{\alpha}{2} = \frac{\sin^2\alpha}{4}$

Similarly for $\beta, \gamma$.

Therefore, the given sum is:

$\frac{\sin^2\alpha + \sin^2\beta + \sin^2\gamma}{4}$

Since $\sin^2\alpha = 1 - \cos^2\alpha = 1 - l^2$, and similarly for $\beta, \gamma$:

$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1 - l^2) + (1 - m^2) + (1 - n^2)$

$= 3 - (l^2 + m^2 + n^2)$

$= 3 - 1 = 2$

Thus, required value = $\frac{2}{4} = \frac{1}{2}$