Derivative of $$\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$$ w.r.t. $$\cos ^{-1} x^2$$ is

A binomial random variable $$\mathrm{X}$$ satisfies $$9. p(X=4)=p(X=2)$$ when $$n=6$$. Then $$p$$ is equal to

If in $$\triangle \mathrm{ABC}$$, with usual notations, $$a \cdot \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$$, then

The lines $$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \quad$$ and $$\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$$