MHTCET 15 Apr 2026 Shift 1 Paper

Concept:For unit vectors, the magnitudes of $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$ can be expressed in terms of $\cos\theta$ using the dot product, then simplified with half-angle identities.Explanation:Since $\bar{a}$ and $\bar{b}$ are unit vectors, we have $|\bar{a}| = 1$ and $|\bar{b}| = 1$.The dot product is $\bar{a}\cdot\bar{b} = |\bar{a}||\bar{b}|\cos\theta = \cos\theta$.Expanding the square of the sum:$$|\bar{a}+\bar{b}|^2 = (\bar{a}+\bar{b})\cdot(\bar{a}+\bar{b}) = |\bar{a}|^2 + |\bar{b}|^2 + 2\bar{a}\cdot\bar{b}$$Substituting the values gives $|\bar{a}+\bar{b}|^2 = 1 + 1 + 2\cos\theta = 2(1+\cos\theta)$.Similarly, expanding the square of the difference:$$|\bar{a}-\bar{b}|^2 = (\bar{a}-\bar{b})\cdot(\bar{a}-\bar{b}) = |\bar{a}|^2 + |\bar{b}|^2 - 2\bar{a}\cdot\bar{b} = 2(1-\cos\theta)$$Taking the ratio of the magnitudes:$$\frac{|\bar{a}+\bar{b}|}{|\bar{a}-\bar{b}|} = \sqrt{\frac{2(1+\cos\theta)}{2(1-\cos\theta)}} = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$$Using the half-angle identities $1+\cos\theta = 2\cos^2\frac{\theta}{2}$ and $1-\cos\theta = 2\sin^2\frac{\theta}{2}$:$$\frac{|\bar{a}+\bar{b}|}{|\bar{a}-\bar{b}|} = \sqrt{\frac{2\cos^2\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}} = \sqrt{\cot^2\frac{\theta}{2}}$$Since $0 < \theta < \pi$, we get $0 < \frac{\theta}{2} < \frac{\pi}{2}$, so $\cot\frac{\theta}{2}$ is positive.Hence, $\frac{|\bar{a}+\bar{b}|}{|\bar{a}-\bar{b}|} = \cot\frac{\theta}{2}$.Answer:Option D: $\cot\frac{\theta}{2}$.