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(A) Let the direction cosines of the line be $l, m, n$.Since the line makes the same angle $	heta$ with the $x$ and $z$-axes,we have $l = \cos 	heta

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$\frac{3}{5}$

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(A) Let the direction cosines of the line be $l, m, n$.Since the line makes the same angle $	heta$ with the $x$ and $z$-axes,we have $l = \cos 	heta$ and $n = \cos 	heta$.The angle with the $y$-axis is $\beta$,so $m = \cos \beta$.The sum of the squares of the direction cosines is always $1$,so $l^2 + m^2 + n^2 = 1$.Substituting the values,we get $\cos^2 	heta + \cos^2 \beta + \cos^2 	heta = 1$,which simplifies to $2 \cos^2 	heta + \cos^2 \beta = 1$.Using the identity $\cos^2 \beta = 1 - \sin^2 \beta$,we have $2 \cos^2 	heta + (1 - \sin^2 \beta) = 1$,which implies $2 \cos^2 	heta = \sin^2 \beta$.Given the condition $\sin^2 \beta = 3 \sin^2 	heta$,we substitute this into our equation: $2 \cos^2 	heta = 3 \sin^2 	heta$.Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get $2 \cos^2 	heta = 3(1 - \cos^2 	heta)$.Expanding this,$2 \cos^2 	heta = 3 - 3 \cos^2 	heta$,which leads to $5 \cos^2 	heta = 3$.Therefore,$\cos^2 	heta = \frac{3}{5}$.

$A$ line makes the same angle $\theta$ with each of the $x$ and $z$-axes. If the angle $\beta$,which it makes with the $y$-axis,is such that $\sin^2 \beta = 3 \sin^2 \theta$,then $\cos^2 \theta$ equals:

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