If 2 sin A = 1, 2 cos B = √2, and π/2 < B < 2π, then find the value of A.

Steps to Solve

1. Understanding the given equations

We start with the equations provided in the problem:

$$ 2\sin A = 1 $$

$$ 2\cos B = \sqrt{2} $$

2. Solving for angle A

From the first equation, we can solve for $\sin A$: $$ \sin A = \frac{1}{2} $$

The sine value corresponds to the angles: $$ A = \frac{\pi}{6} \quad \text{or} \quad A = \frac{5\pi}{6} $$

3. Solving for angle B

From the second equation, we solve for $\cos B$: $$ \cos B = \frac{\sqrt{2}}{2} $$

This corresponds to the angles: $$ B = \frac{\pi}{4} \quad \text{or} \quad B = \frac{7\pi}{4} $$

4. Verifying the range for B

We know the condition is: $$ \frac{3\pi}{2} < B < 2\pi $$

So the applicable value for $B$ is: $$ B = \frac{7\pi}{4} $$

5. Finding the value of $\tan A + \tan B$ and $\frac{\tan A + \tan B}{\cos A - \cos B}$

Calculate $\tan A$ and $\tan B$:

For $A = \frac{\pi}{6}$: $$ \tan A = \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} $$

For $B = \frac{7\pi}{4}$: $$ \tan B = \tan \left( \frac{7\pi}{4} \right) = -1 $$

Then calculate: $$ \tan A + \tan B = \frac{1}{\sqrt{3}} - 1 $$

6. Finding $\cos A$ and $\cos B$

Calculate $\cos A$ and $\cos B$:

For $A = \frac{\pi}{6}$: $$ \cos A = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} $$

For $B = \frac{7\pi}{4}$: $$ \cos B = \cos \left( \frac{7\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

Now calculate: $$ \cos A - \cos B = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} $$

7. Final expression

Putting it all together:

$$ \frac{\tan A + \tan B}{\cos A - \cos B} = \frac{\frac{1}{\sqrt{3}} - 1}{\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}} $$