If tan(pi/4 + theta) + tan(pi/4 − theta) = 3, then the value of tan²(pi/4 + theta) + tan²(pi/4 - theta) is?

We are given the equation:

$$x + y = 3$$

We need to find the value of $x^2 + y^2$.
We know that $x^2 + y^2 = (x+y)^2 - 2xy$.

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:

$$x = \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan \theta}{1 - \tan\left(\frac{\pi}{4}\right) \tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta}$$ $$y = \tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan \theta}{1 + \tan\left(\frac{\pi}{4}\right) \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}$$

$$x^2 + y^2 = (x+y)^2 - 2xy$$ $$x^2 + y^2 = (3)^2 - 2(1)$$ $$x^2 + y^2 = 9 - 2$$ $$x^2 + y^2 = 7$$

The value of $\tan^2\left(\frac{\pi}{4} + \theta\right) + \tan^2\left(\frac{\pi}{4} - \theta\right)$ is $\boxed{\text{7}}$.

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Question 7:

We are given:

$$2 \tan A = 3 \tan B$$

We need to find $\tan(A-B)$.
The formula for $\tan(A-B)$ is:

$$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

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:

$$\tan(A-B) = \frac{\tan B}{2 + 3 \tan^2 B}$$

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 $\boxed{\frac{\sin 2B}{5 - \cos 2B}}$.