JEE Main 2025 Jan 23 Shift 1, Mathematics Q14: Trigonometric Ratios And Identities

Pick the auxiliary angle $\alpha$ with $\cos \alpha =\frac{12}{13}$ and $\sin \alpha =\frac{5}{13}$, so $\alpha ={\tan }^{-1}\frac{5}{12}$, and $\alpha \in (0,\pi /2)$.

Then $\frac{12}{13}\cos x+\frac{5}{13}\sin x=\cos \alpha \cos x+\sin \alpha \sin x=\cos (x-\alpha )$.

For $x\in [\pi /2,3\pi /4]$, $x-\alpha \in [\pi /2-\alpha ,3\pi /4-\alpha ]\subset (0,\pi )$ where ${\cos }^{-1}(\cos \theta )=\theta$.

$\therefore {\cos }^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)=x-\alpha =x-{\tan }^{-1}\frac{5}{12}$.