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(D) We know that for direction cosines,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Since $\sin^2 	heta = 1 - \cos^2 	heta$,we have $\sin^2 \

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$\frac{\pi}{3}$

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(D) We know that for direction cosines,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Since $\sin^2 	heta = 1 - \cos^2 	heta$,we have $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) = 3 - 1 = 2$. The given equation is $2(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 3 \sec^2 \frac{	heta}{2}$. Substituting the value,$2(2) = 3 \sec^2 \frac{	heta}{2} \Rightarrow 4 = 3 \sec^2 \frac{	heta}{2}$. $\sec^2 \frac{	heta}{2} = \frac{4}{3} \Rightarrow \cos^2 \frac{	heta}{2} = \frac{3}{4}$. Using the identity $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,we get $\cos 	heta = 2(\frac{3}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}$. Thus,$	heta = \frac{\pi}{3}$.

If $\alpha, \beta, \gamma$ are the angles made by a line with $x, y,$ and $z$ axes respectively such that $2\left( \frac{\tan^2 \alpha}{1 + \tan^2 \alpha} + \frac{\tan^2 \beta}{1 + \tan^2 \beta} + \frac{\tan^2 \gamma}{1 + \tan^2 \gamma} \right) = 3 \sec^2 \frac{\theta}{2}$,then $\theta =$

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