Step 1: Let $x = \tan\left(\frac{\pi}{4} + \theta\right)$ and $y = \tan\left(\frac{\pi}{4} - \theta\right)$. We are given the equation: $$x + y = 3$$ We need to find the value of $x^2 + y^2$. We know that $x^2 + y^2 = (x+y)^2 - 2xy$. Step 2: Use the tangent addition and subtraction formulas. The formula for $\tan(A+B)$ is $\frac{\tan A + \tan B}{1 - \tan A \tan B}$. The formula for $\tan(A-B)$ is $\frac{\tan A - \tan B}{1 + \tan A \tan B}$. Here, $A = \frac{\pi}{4}$, so $\tan A = \tan\left(\frac{\pi}{4}\right) = 1$. So, we have: $$x = \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan \theta}{1 - \tan\left(\frac{\pi}{4}\right) \tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta}$$ $$y = \tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan \theta}{1 + \tan\left(\frac{\pi}{4}\right) \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}$$ Step 3: Calculate the product $xy$. $$xy = \left(\frac{1 + \tan \theta}{1 - \tan \theta}\right) \left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) = 1$$ Step 4: Substitute the values of $(x+y)$ and $xy$ into the expression for $x^2 + y^2$. $$x^2 + y^2 = (x+y)^2 - 2xy$$ $$x^2 + y^2 = (3)^2 - 2(1)$$ $$x^2 + y^2 = 9 - 2$$ $$x^2 + y^2 = 7$$ The value of $\tan^2\left(\frac{\pi}{4} + \theta\right) + \tan^2\left(\frac{\pi}{4} - \theta\right)$ is $\boxed{\text{7}}$. --- Question 7: Step 1: Write down the given information and the formula for $\tan(A-B)$. We are given: $$2 \tan A = 3 \tan B$$ We need to find $\tan(A-B)$. The formula for $\tan(A-B)$ is: $$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$ Step 2: Express $\tan A$ in terms of $\tan B$ and substitute it into the formula. From the given equation, we have $\tan A = \frac{3}{2} \tan B$. Substitute this into the $\tan(A-B)$ formula: $$\tan(A-B) = \frac{\frac{3}{2} \tan B - \tan B}{1 + \left(\frac{3}{2} \tan B\right) \tan B}$$ $$\tan(A-B) = \frac{\left(\frac{3}{2} - 1\right) \tan B}{1 + \frac{3}{2} \tan^2 B}$$ $$\tan(A-B) = \frac{\frac{1}{2} \tan B}{1 + \frac{3}{2} \tan^2 B}$$ Multiply the numerator and denominator by 2 to simplify: $$\tan(A-B) = \frac{\tan B}{2 + 3 \tan^2 B}$$ Step 3: Compare the result with the given options by expressing the options in terms of $\tan B$. Let $t = \tan B$. Then $\tan(A-B) = \frac{t}{2 + 3t^2}$. We use the double angle formulas: $$\sin 2B = \frac{2 \tan B}{1 + \tan^2 B} = \frac{2t}{1+t^2}$$ $$\cos 2B = \frac{1 - \tan^2 B}{1 + \tan^2 B} = \frac{1-t^2}{1+t^2}$$ Let's check option 2: $\frac{\sin 2B}{5 - \cos 2B}$. Substitute the expressions for $\sin 2B$ and $\cos 2B$: $$\frac{\frac{2t}{1+t^2}}{5 - \frac{1-t^2}{1+t^2}} = \frac{\frac{2t}{1+t^2}}{\frac{5(1+t^2) - (1-t^2)}{1+t^2}}$$ $$= \frac{2t}{5(1+t^2) - (1-t^2)}$$ $$= \frac{2t}{5 + 5t^2 - 1 + t^2}$$ $$= \frac{2t}{4 + 6t^2}$$ $$= \frac{t}{2 + 3t^2}$$ This matches our derived expression for $\tan(A-B)$. The value of $\tan(A-B)$ is $\boxed{\frac{\sin 2B}{5 - \cos 2B}}$.