If a line makes angles tan ^{-1} sqrt{7} and | vedclass

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(D) We know that if a line makes angles $\alpha, \beta, \gamma$ with the positive direction of $X$-axis,$Y$-axis,and $Z$-axis respectively,then $\cos^

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$\frac{\pi}{4}$ or $\frac{3 \pi}{4}$

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(D) We know that if a line makes angles $\alpha, \beta, \gamma$ with the positive direction of $X$-axis,$Y$-axis,and $Z$-axis respectively,then $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given $\alpha = 	an^{-1} \sqrt{7}$ and $\beta = 	an^{-1} \sqrt{\frac{5}{3}}$. For $\alpha = 	an^{-1} \sqrt{7}$,we have $\cos \alpha = \frac{1}{\sqrt{1 + (\sqrt{7})^2}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$. For $\beta = 	an^{-1} \sqrt{\frac{5}{3}}$,we have $\cos \beta = \frac{1}{\sqrt{1 + (\sqrt{5/3})^2}} = \frac{1}{\sqrt{1 + 5/3}} = \frac{1}{\sqrt{8/3}} = \sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2\sqrt{2}}$. Substituting these values into the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$: $\left(\frac{1}{2\sqrt{2}}ight)^2 + \left(\frac{\sqrt{3}}{2\sqrt{2}}ight)^2 + \cos^2 \gamma = 1$ $\frac{1}{8} + \frac{3}{8} + \cos^2 \gamma = 1$ $\frac{4}{8} + \cos^2 \gamma = 1$ $\frac{1}{2} + \cos^2 \gamma = 1$ $\cos^2 \gamma = \frac{1}{2}$ $\cos \gamma = \pm \frac{1}{\sqrt{2}}$ Therefore,$\gamma = \frac{\pi}{4}$ or $\gamma = \frac{3\pi}{4}$.

If a line makes angles $\tan ^{-1} \sqrt{7}$ and $\tan ^{-1}\left(\sqrt{\frac{5}{3}}\right)$ with $X$-axis and $Y$-axis respectively,then the angle made by it with $Z$-axis is

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