AP DSC School Assistant Math Model Test 3

Concept:Use the identity $\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta$ to rewrite the equation, then apply the condition $\tan^{-1}A + \tan^{-1}B = \frac{\pi}{2}$ which implies $AB = 1$.Explanation:Given: $\tan^{-1}\left(\frac{2}{3^{-x}+1}\right) = \cot^{-1}\left(\frac{3}{3^{x}+1}\right)$.Let $A = \frac{2}{3^{-x}+1}$ and $B = \frac{3}{3^{x}+1}$. Using the identity:$\cot^{-1}B = \frac{\pi}{2} - \tan^{-1}B$. So the equation becomes:$\tan^{-1}A = \frac{\pi}{2} - \tan^{-1}B$ → $\tan^{-1}A + \tan^{-1}B = \frac{\pi}{2}$.Taking tangent of both sides: $\tan(\tan^{-1}A + \tan^{-1}B) = \tan\frac{\pi}{2}$ which is undefined, so the denominator $1-AB$ must be zero:$1 - AB = 0$ → $AB = 1$.Thus $\frac{2}{3^{-x}+1} \cdot \frac{3}{3^{x}+1} = 1$.Simplify: $6 = (3^{-x}+1)(3^{x}+1)$. Expand the product:$(3^{-x})(3^{x}) + 3^{-x} + 3^{x} + 1 = 1 + 3^{-x} + 3^{x} + 1 = 3^{x} + 3^{-x} + 2$.So $6 = 3^{x} + 3^{-x} + 2$ → $3^{x} + 3^{-x} = 4$.Let $y = 3^{x} > 0$. Then $3^{-x} = \frac{1}{y}$. The equation becomes $y + \frac{1}{y} = 4$. Multiply by $y$:$y^{2} - 4y + 1 = 0$. Solve:\(y = \frac{