JEE Main 2025 Jan 29 Shift 2, Mathematics Q2: Trigonometric Ratios And Identities

The hypothesis $\sin x=1-{\sin }^{2}x={\cos }^{2}x$ implies $\tan x=\frac{\sin x}{\cos x}=\frac{{\cos }^{2}x}{\cos x}=\cos x$.

Substitute $\tan x=\cos x$ everywhere:

$E=2{\cos }^{12}x+6{\cos }^{10}x+6{\cos }^{8}x+2{\cos }^{6}x=2{\cos }^{6}x\left({\cos }^{6}x+3{\cos }^{4}x+3{\cos }^{2}x+1\right)$

The bracketed expression is $({\cos }^{2}x+1{)}^3$. With ${\cos }^{2}x=\sin x$,

$E=2{\sin }^{3}x(\sin x+1{)}^3=2[\sin x(\sin x+1){]}^3=2({\sin }^{2}x+\sin x{)}^3=2\cdot 1^3=2$.