If \(\cot\theta+\tan\theta=2\sec\theta\), where \(0^\circ<\theta<90^\circ\), what is the value of \(\displaystyle\frac{\tan\theta-\sec\theta}{2\tan\theta+3\sec\theta}\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{1}{8}\)
Writing everything in sine and cosine, \(\cot\theta+\tan\theta=\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}\). The equation then fixes \(\sin\theta\), and the acute angle follows.
\(\displaystyle\frac{1}{\sin\theta\cos\theta}=\frac{2}{\cos\theta}\), so \(\sin\theta=\frac{1}{2}\) and \(\theta=30^\circ\)
\(\tan 30^\circ=\frac{1}{\sqrt{3}}\) and \(\sec 30^\circ=\frac{2}{\sqrt{3}}\)
Numerator \(\displaystyle=\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\)
Denominator \(\displaystyle=\frac{2}{\sqrt{3}}+\frac{6}{\sqrt{3}}=\frac{8}{\sqrt{3}}\)
Value \(\displaystyle=-\frac{1}{8}\)