Two vectors $$\vec{P}$$ and $$\vec{Q}$$ have equal magnitudes. If the magnitude of $$\vec{P} + \vec{Q}$$ is $$n$$ times the magnitude of $$\vec{P} - \vec{Q}$$, then angle between $$\vec{P}$$ and $$\vec{Q}$$ is:

Let the magnitude of each vector be $$P$$, and let $$\theta$$ be the angle between $$\vec{P}$$ and $$\vec{Q}$$.

The magnitude of $$\vec{P} + \vec{Q}$$ is $$|\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$. Since $$P = Q$$, this becomes $$\sqrt{2P^2 + 2P^2\cos\theta} = P\sqrt{2(1+\cos\theta)}$$.

The magnitude of $$\vec{P} - \vec{Q}$$ is $$|\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2 - 2PQ\cos\theta} = P\sqrt{2(1-\cos\theta)}$$.

Setting $$|\vec{P} + \vec{Q}| = n|\vec{P} - \vec{Q}|$$ gives $$P\sqrt{2(1+\cos\theta)} = nP\sqrt{2(1-\cos\theta)}$$, so $$1+\cos\theta = n^2(1-\cos\theta)$$.

Solving: $$1 + \cos\theta = n^2 - n^2\cos\theta$$, hence $$\cos\theta(1 + n^2) = n^2 - 1$$, giving $$\cos\theta = \dfrac{n^2 - 1}{n^2 + 1}$$.

Therefore the angle between the two vectors is $$\theta = \cos^{-1}\!\left(\dfrac{n^2-1}{n^2+1}\right)$$.