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A line makes the same angle $\theta$ with each of the $\mathrm{X}$ and $\mathrm{Z}$ -axes. If the angle $\beta$, which it makes with Y-axis, is such that $\sin ^{2} \beta=3 \sin ^{2} \theta,$ then $\cos ^{2} \theta$ equals
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The correct answer is:
$3 / 5$
Let $l, \mathrm{~m}$ and $\mathrm{n}$ be the direction cosines. Then, $l=\cos \theta, \mathrm{m}=\cos \beta, \mathrm{n}=\cos \theta$
$$
\begin{array}{l}
\text { we have } l^{2}+m^{2}+n^{2}=1 \\
\Rightarrow \cos ^{2} \theta+\cos ^{2} \beta+\cos ^{2} \theta=1 \\
\Rightarrow 2 \cos ^{2} \theta+1-\sin ^{2} \beta=1 \\
\Rightarrow 2 \cos ^{2} \theta-\sin ^{2} \beta=0 \\
\Rightarrow 2 \cos ^{2} \theta-3 \sin ^{2} \beta=0 \\
\quad\left[\because \sin ^{2} \beta=3 \sin ^{2} 0(\text { given })\right] \\
\Rightarrow \tan ^{2} \theta=2 / 3 \\
\therefore \cos ^{2} \theta=\frac{1}{1+\tan ^{2} \theta}=\frac{1}{1+2 / 3}=\frac{3}{5}
\end{array}
$$
$$
\begin{array}{l}
\text { we have } l^{2}+m^{2}+n^{2}=1 \\
\Rightarrow \cos ^{2} \theta+\cos ^{2} \beta+\cos ^{2} \theta=1 \\
\Rightarrow 2 \cos ^{2} \theta+1-\sin ^{2} \beta=1 \\
\Rightarrow 2 \cos ^{2} \theta-\sin ^{2} \beta=0 \\
\Rightarrow 2 \cos ^{2} \theta-3 \sin ^{2} \beta=0 \\
\quad\left[\because \sin ^{2} \beta=3 \sin ^{2} 0(\text { given })\right] \\
\Rightarrow \tan ^{2} \theta=2 / 3 \\
\therefore \cos ^{2} \theta=\frac{1}{1+\tan ^{2} \theta}=\frac{1}{1+2 / 3}=\frac{3}{5}
\end{array}
$$
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