Question

If $\alpha$ and $\beta$ are acute angles such that $\alpha + \beta$ and $\alpha - \beta$ satisfy the equation $\tan^2 \theta - 4\tan \theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively

• A. $45^\circ, 30^\circ$
• B. $75^\circ, 15^\circ$
• C. $30^\circ, 60^\circ$
• D. $60^\circ, 45^\circ$

Hint:

Memorizing the trigonometric values for non-standard but common angles like $15^\circ$ and $75^\circ$ ($\tan 15^\circ = 2 - \sqrt{3}$, $\tan 75^\circ = 2 + \sqrt{3}$) can significantly speed up solving such problems.

Answer 1

The Correct Option is A

Step 1: Form the quadratic equation
Let $x = \tan \theta$. Then, \[ x^2 - 4x + 1 = 0 \]
Step 2: Solve the quadratic equation
\[ x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Step 3: Identify the angles
\[ \tan(\alpha + \beta) = 2 + \sqrt{3} = \tan 75^\circ \] \[ \tan(\alpha - \beta) = 2 - \sqrt{3} = \tan 15^\circ \]
Step 4: Form equations
\[ \alpha + \beta = 75^\circ (1) \] \[ \alpha - \beta = 15^\circ (2) \]
Step 5: Solve for $\alpha$ and $\beta$
Adding (1) and (2): \[ 2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ \] Subtracting (2) from (1): \[ 2\beta = 60^\circ \Rightarrow \beta = 30^\circ \] Final Answer:
\[ \alpha = 45^\circ, \beta = 30^\circ \]