Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\frac{k+1}{k-1}=\frac{\sin 4A}{\sin 2A}=2\cos 2A$, so $\frac{\sin 3A}{\sin A}=1+2\cos 2A=\frac{2k}{k-1}$ and $\frac{\cos 3A}{\cos A}=2\cos 2A-1=\frac{2}{k-1}$. Also $k=\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}\Rightarrow k<\frac{1}{3}$ or $k>3$.

Since $k\in (-\infty ,\frac{1}{3})\cup (3,\infty )$, statements (c) and (d) each hold only for some values of A, not for every A. So only (a) and (b) always hold: correct answer A, B.