Question

The value of $\tan^{-1}\left(\frac{\cos x-\sqrt{3}\sin x}{\sqrt{3}\cos x+\sin x}\right)$ , where $0<x<\frac{\pi}{2}$ is

• A. $\frac{\pi}{6}-x$
• B. $\frac{\pi}{4}-x$
• C. $\frac{\pi}{3}-x$
• D. $\frac{\pi}{2}-x$
• E. $\pi-x$

Hint:

Logic Tip: Whenever you see an expression of the form $\frac{a\cos x - b\sin x}{b\cos x + a\sin x}$ inside an inverse tangent, immediately divide everything by $b\cos x$ (the bottom-left term) to create the $1 + \dots$ structure needed for the $\tan(A-B)$ formula.

Answer 1

The Correct Option is A

Concept:
The goal is to convert the expression inside the inverse tangent into the form $\tan(A - B)$ so that the $\tan$ and $\tan^{-1}$ cancel each other. The compound angle formula is: $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Step 1: Divide the numerator and denominator by $\sqrt{3}\cos x$.
To force the expression into the format of the tangent difference formula, we divide every term inside the parentheses by $\sqrt{3}\cos x$: $$\frac{\frac{\cos x}{\sqrt{3}\cos x} - \frac{\sqrt{3}\sin x}{\sqrt{3}\cos x}}{\frac{\sqrt{3}\cos x}{\sqrt{3}\cos x} + \frac{\sin x}{\sqrt{3}\cos x}}$$
Step 2: Simplify the fractions into tangents.
$$= \frac{\frac{1}{\sqrt{3}} - \tan x}{1 + \frac{1}{\sqrt{3}}\tan x}$$
Step 3: Identify the standard trigonometric value.
We know that $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. Substitute this into the expression: $$= \frac{\tan\left(\frac{\pi}{6}\right) - \tan x}{1 + \tan\left(\frac{\pi}{6}\right)\tan x}$$
Step 4: Apply the tangent difference formula and evaluate.
The structure perfectly matches the formula for $\tan(A - B)$, where $A = \frac{\pi}{6}$ and $B = x$: $$= \tan\left(\frac{\pi}{6} - x\right)$$ Now return to the original inverse tangent function: $$\tan^{-1}\left( \tan\left(\frac{\pi}{6} - x\right) \right)$$ Since $0<x<\frac{\pi}{2}$, the angle $\frac{\pi}{6} - x$ lies within the principal range of $\tan^{-1}$ $(-\frac{\pi}{2}, \frac{\pi}{2})$, allowing the functions to cancel: $$= \frac{\pi}{6} - x$$