Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Given, $3{\cos }^2\mathrm{A}+2{\cos }^2\mathrm{B}=4$ $\Rightarrow 2{\cos }^2\mathrm{B}-1=4-3{\cos }^2\mathrm{A}-1$ $\Rightarrow \cos 2\mathrm{B}=3\left(1-{\cos }^2\mathrm{A}\right)=3{\sin }^2\mathrm{A}\dots (1)$ and $2\cos \mathrm{B}\sin \mathrm{B}=3\sin \mathrm{A}\cos \mathrm{A}$ $\sin 2\mathrm{B}=3\sin \mathrm{A}\cos \mathrm{A}$ Now, $\cos (\mathrm{A}+2\mathrm{B})=\cos \mathrm{A}\cos 2\mathrm{B}-\sin \mathrm{A}$ $\sin 2\mathrm{B}$ $=\cos A\left(3{\sin }^{2}A\right)-\sin A(3\sin A\cos A)=0$ [using eqs. (1) and (2)] $\Rightarrow A+2B=\frac{\pi }{2}$