Prove the following


Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $1+m-n=0$ and $\mathrm{I}^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0$. Then the value of $\sin ^{4} \alpha+\cos ^{4} \alpha$ is :

  1. (1) $\frac{3}{4}$

  2. (2) $\frac{1}{2}$

  3. (3) $\frac{5}{8}$

  4. (4) $\frac{3}{8}$

Correct Option: , 3



$\therefore 2 n^{2}=1 \Rightarrow n=\pm \frac{1}{\sqrt{2}}$

$\left.\therefore\right|^{2}+m^{2}=\frac{1}{2} \& \mid+m=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{2}-2 \operatorname{lm}=\frac{1}{2}$

$\Rightarrow 1 \mathrm{~m}=0$ or $\mathrm{m}=0$

$<0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}>\quad$ or $<\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}>$

$\therefore \cos \alpha=0+0+\frac{1}{2}=\frac{1}{2}$

$\therefore \sin ^{4} \alpha+\cos ^{4} \alpha=1-\frac{1}{2} \sin ^{2}(2 \alpha)=1-\frac{1}{2}, \frac{3}{4}=\frac{5}{8}$

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