If \(\displaystyle\frac{\sin x}{\sin y}=\frac{1}{2}\) and \(\displaystyle\frac{\cos x}{\cos y}=\frac{3}{2}\), then find the value of \(\cot^2 x+\sin^2 x\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{889}{160}\)
Both ratios give \(\sin y\) and \(\cos y\) in terms of \(x\). Putting them into \(\sin^2 y+\cos^2 y=1\) removes \(y\) and leaves an equation for \(\sin^2 x\).
\(\sin y=2\sin x\) and \(\cos y=\displaystyle\frac{2}{3}\cos x\)
\(4\sin^2 x+\displaystyle\frac{4}{9}\cos^2 x=1\); multiplying by 9: \(36\sin^2 x+4(1-\sin^2 x)=9\)
\(32\sin^2 x=5\), so \(\sin^2 x=\displaystyle\frac{5}{32}\) and \(\cos^2 x=\displaystyle\frac{27}{32}\)
\(\cot^2 x\displaystyle=\frac{\cos^2 x}{\sin^2 x}=\frac{27}{5}\)
\(\cot^2 x+\sin^2 x\displaystyle=\frac{27}{5}+\frac{5}{32}=\frac{864+25}{160}=\frac{889}{160}\)