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(B) Since the line makes an equal angle $	heta$ with the three coordinate axes,we have $\alpha = \beta = \gamma = 	heta$. Therefore,the direction co

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$\sqrt{2}$

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(B) Since the line makes an equal angle $	heta$ with the three coordinate axes,we have $\alpha = \beta = \gamma = 	heta$. Therefore,the direction cosines are $l = m = n = \cos 	heta$. We know that for any line,$l^2 + m^2 + n^2 = 1$. Substituting the values,we get $3 \cos^2 	heta = 1$,which implies $\cos^2 	heta = \frac{1}{3}$. Thus,$\cos 	heta = \frac{1}{\sqrt{3}}$. Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get $\sin^2 	heta = 1 - \frac{1}{3} = \frac{2}{3}$,so $\sin 	heta = \frac{\sqrt{2}}{\sqrt{3}}$. Finally,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{2}/\sqrt{3}}{1/\sqrt{3}} = \sqrt{2}$.

If a straight line is equally inclined at an angle $\theta$ with all the three coordinate axes,then $\tan \theta=$

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