JEE Main 2025 Apr 4 Shift 2, Mathematics Q5: Trigonometric Ratios And Identities

Group real and imaginary parts.

${\omega }_1=(8\sin \theta +7\cos \theta )+i(\sin \theta +4\cos \theta )$,

${\omega }_2=(\sin \theta +4\cos \theta )+i(8\sin \theta +7\cos \theta )$.

Write ${\omega }_1=A+iB$ and ${\omega }_2=B+iA$, where $A=8\sin \theta +7\cos \theta$ and $B=\sin \theta +4\cos \theta$. Then

${\omega }_1{\omega }_2=(A+iB)(B+iA)=(AB-AB)+i(A^2+B^2)=i(A^2+B^2)$.

So $\alpha =0$ and $\beta =A^2+B^2$. Compute:

$A^2+B^2=(8\sin \theta +7\cos \theta {)}^2+(\sin \theta +4\cos \theta {)}^2$

$=65{\sin }^2\theta +65{\cos }^2\theta +2(56+4)\sin \theta \cos \theta =65+60\sin 2\theta$.

Hence $\alpha +\beta =65+60\sin 2\theta$, with maximum $p=125$ and minimum $q=5$.

$\therefore p+q=130$.