Discussion
The value of \(tan^{-1}\left(\frac{cosx - \sqrt{3}sinx}{\sqrt{3}cosx + sinx}\right)\) , where \(0 \lt x \lt \frac{\pi}{2}\) is
Answer: A
The value is \(\frac{\pi}{6} - x\).
Divide the numerator and denominator by 2 and recognise the compound-angle forms.
\[ \begin{aligned} \frac{\tfrac{1}{2}\cos x - \tfrac{\sqrt{3}}{2}\sin x}{\tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x} &= \frac{\cos(x + \frac{\pi}{3})}{\sin(x + \frac{\pi}{3})} \\ &= \tan\left(\tfrac{\pi}{6} - x\right) \end{aligned} \]
For \(0 \lt x \lt \frac{\pi}{2}\), \(\frac{\pi}{6} - x\) lies in \((-\frac{\pi}{3}, \frac{\pi}{6})\), inside the principal range of \(\tan^{-1}\), so \(\tan^{-1}\) returns it unchanged. \(\frac{\pi}{3} - x\) uses the angle \(\frac{\pi}{3}\) without converting the cotangent into a tangent.
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