Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
If $u=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}+\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$, then the difference between maximum and minimum values of $u^2$ is given by:
(a-b)^2
(a+b)^2
a^2+b^2
None of these
Step-by-Step Solution
Key Concept: Critical values of trigonometric expressions occur at extrema of sine and cosine functions.
We simplify $u^2 = (a^2 + b^2) + 2|a^2b^2(1-2\sin^2\theta\cos^2\theta) + \sin^2\theta\cos^2\theta(a^4 + b^4)|^{1/2}$ to get $u^2 = (a^2 + b^2) + |4u^2b^2 + \sin^2 2\theta(a^2-b^2)|^{1/2}$. The maximum occurs when $\sin 2\theta = 0$, giving $u^2_{max} = 2(a^2+b^2)$, and minimum when $\sin 2\theta = 1$, giving $u^2_{min} = a^2 + b^2 + 2ab$. The difference is $(a^2+b^2) - (a^2+b^2+2ab) = (a-b)^2$.
Correct Answer: 1