JEE Main 2019 (Online) 9th January Morning ShiftQuestion 1
For any $\theta\in\left(\frac\pi4,\frac\pi2\right)$, the expression $3(\sin\theta-\cos\theta)^4+6(\sin\theta+\cos\theta)^2+4\sin^6\theta$ equals:
Trigonometric Ratio & Identities
Solution
Concept. The first two terms give $9+3\sin^22\theta$; then write everything with $C=\cos^2\theta$. Formulas used. (1) $(\sin\theta\pm\cos\theta)^2=1\pm\sin2\theta$, $\sin^22\theta=4\sin^2\theta\cos^2\theta$Step 1 - First two terms. By (1), $$3(1-\sin2\theta)^2+6(1+\sin2\theta)=9+3\sin^22\theta=9+12\sin^2\theta\cos^2\theta.$$ Step 2 - Use $C=\cos^2\theta$. With $\sin^2\theta=1-C$: $$9+12C(1-C)+4(1-C)^3=9+12C-12C^2+4-12C+12C^2-4C^3=13-4C^3.$$ Step 3 - Value. $13-4\cos^6\theta$. Answer: (C)