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

The Correct Option is A

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:
\(A + B = C\)