Question

If $\tan A = \frac{1}{\sqrt{x(x^2 + x + 1)}}, \quad \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}$ and $\tan C = \left(x^3 + x^2 + x^{-1}\right)^{\frac{1}{2}}, \quad 0 < A, B, C < \frac{\pi}{2}$,then $A + B$ is equal to:

• A. $C$
• B. $\pi - C$
• C. $2\pi - C$
• D. $\frac{\pi}{2} - C$

Answer 1

Answer: (A)

$C$

Answer 2

Finding $\tan(A + B)$, we get:

$\implies \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
Substituting the values:

$\implies \tan(A + B) = \frac{\frac{1}{x^2+x+1} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \frac{\sqrt{x}}{x^2+x+1}}$
Simplifying the numerator and denominator:

$\implies \tan(A + B) = \frac{(1 + x)(\sqrt{x^2+x+1})}{(x^2+x)(\sqrt{x})}$

Rewriting:

$\tan(A + B) = \frac{\sqrt{x^2 + x + 1}}{x\sqrt{x}} = \tan C$
Therefore:
$A + B = C$