Question

If A & B are acute angles such the A + B and A – B satisfy the equation $tan^2 \theta - 4 \tan \theta + 1 = 0$ then

• A. A = $\pi/4$
• B. A = $\pi/6$
• C. B = $\pi/4$
• D. B = $\pi/6$

Answer 1

Answer: (A), (D)

A = $\pi/4$ and B = $\pi/6$

Answer 2

Let $tan(A+B)$ and $tan(A-B)$ be the roots of the quadratic equation $tan^2 \theta - 4 \tan \theta + 1 = 0$.

From Vieta's formulas: Sum of roots: $tan(A+B) + tan(A-B) = 4$ Product of roots: $tan(A+B) \cdot tan(A-B) = 1$

Using the product of roots: $\frac{tan A + tan B}{1 - tan A tan B} \cdot \frac{tan A - tan B}{1 + tan A tan B} = 1$ $\frac{tan^2 A - tan^2 B}{1 - tan^2 A tan^2 B} = 1$ $tan^2 A - tan^2 B = 1 - tan^2 A tan^2 B$ $tan^2 A + tan^2 A tan^2 B = 1 + tan^2 B$ $tan^2 A (1 + tan^2 B) = 1 + tan^2 B$

Since A and B are acute angles, $tan B > 0$, so $1 + tan^2 B \neq 0$. Dividing by $(1 + tan^2 B)$, we get: $tan^2 A = 1$ Since A is an acute angle, $tan A > 0$, so $tan A = 1$. This implies $A = \pi/4$.

Now, substitute $A = \pi/4$ into the sum of roots equation: $tan(A+B) + tan(A-B) = 4$ $tan(\pi/4 + B) + tan(\pi/4 - B) = 4$ Using the tangent addition and subtraction formulas: $\frac{tan(\pi/4) + tan B}{1 - tan(\pi/4) tan B} + \frac{tan(\pi/4) - tan B}{1 + tan(\pi/4) tan B} = 4$ $\frac{1 + tan B}{1 - tan B} + \frac{1 - tan B}{1 + tan B} = 4$ Let $t = tan B$. $\frac{(1+t)^2 + (1-t)^2}{(1-t)(1+t)} = 4$ $\frac{(1 + 2t + t^2) + (1 - 2t + t^2)}{1 - t^2} = 4$ $\frac{2 + 2t^2}{1 - t^2} = 4$ $2(1 + t^2) = 4(1 - t^2)$ $1 + t^2 = 2(1 - t^2)$ $1 + t^2 = 2 - 2t^2$ $3t^2 = 1$ $t^2 = 1/3$ $tan^2 B = 1/3$ Since B is an acute angle, $tan B > 0$, so $tan B = 1/\sqrt{3}$. This implies $B = \pi/6$.

Thus, $A = \pi/4$ and $B = \pi/6$.