Trigonometric Ratios & Identities
Addition of Angles / tan(A+B)
nta_pyq_2024_jan
Grade 11
Question:
Let $\tan A = \dfrac{1}{\sqrt{x(x^2+x+1)}}$, $\tan B = \dfrac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C = \sqrt{x^{-3}+x^{-2}+x^{-1}}$, where $0 < A, B, C < \dfrac{\pi}{2}$. Then $A+B$ is equal to:
$C$
$\pi - C$
$2\pi - C$
$\dfrac{\pi}{2}$
Step-by-Step Solution
Key Concept: Compute $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Note $\tan A\cdot\tan B=\dfrac{1}{x^2+x+1}$, so $1-\tan A\tan B=\dfrac{x^2+x}{x^2+x+1}$. Numerator simplifies to $\dfrac{1+x}{\sqrt{x}\cdot\sqrt{x^2+x+1}}$. After division: $\tan(A+B)=\dfrac{\sqrt{x^2+x+1}}{x\sqrt{x}}=\sqrt{\dfrac{x^2+x+1}{x^3}}=\tan C$.
$\tan(A+B)=\dfrac{\frac{1}{\sqrt{x(x^2+x+1)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}}=\dfrac{\frac{(1+x)\sqrt{x^2+x+1}}{(x^2+x)\sqrt{x}}}{1}=\dfrac{\sqrt{x^2+x+1}}{x\sqrt{x}}=\tan C$. Since all angles are in $(0,\frac{\pi}{2})$, $A+B=C$.
Correct Answer: 1