Finding the Ratio tan A : tan B from Trigonometric Equations

This problem asks us to find the ratio of tan A to tan B, given two equations involving trigonometric functions of sums and differences of angles A and B, along with constraints on the values of A and B.

The given equations are:

1. sin(A + B) = 1
2. 2 sin(A - B) = 1

We are also given that 0 < A < \(\frac{\pi}{2}\) and 0 < B < \(\frac{\pi}{2}\).

Step 1: Solve the First Equation for (A + B)

The first equation is sin(A + B) = 1.

We know that the sine function equals 1 for angles of the form \(\frac{\pi}{2} + 2n\pi\), where n is an integer.

Given the constraints 0 < A < \(\frac{\pi}{2}\) and 0 < B < \(\frac{\pi}{2}\), the sum A + B must satisfy 0 < A + B < \(\pi\).

Within the interval (0, \(\pi\)), the only angle whose sine is 1 is \(\frac{\pi}{2}\).

Therefore, from sin(A + B) = 1, we conclude:

A + B = \(\frac{\pi}{2}\) (Equation 1)

Step 2: Solve the Second Equation for (A - B)

The second equation is 2 sin(A - B) = 1.

Dividing by 2, we get sin(A - B) = \(\frac{1}{2}\).

The sine function equals \(\frac{1}{2}\) for angles of the form \(\frac{\pi}{6} + 2n\pi\) or \(\frac{5\pi}{6} + 2n\pi\), where n is an integer.

Now consider the constraints on A - B. Since 0 < A < \(\frac{\pi}{2}\) and 0 < B < \(\frac{\pi}{2}\), the difference A - B must satisfy \(0 - \frac{\pi}{2} < A - B < \frac{\pi}{2} - 0\), which simplifies to \(-\frac{\pi}{2} < A - B < \frac{\pi}{2}\).

Within the interval (\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)), the only angle whose sine is \(\frac{1}{2}\) is \(\frac{\pi}{6}\).

Therefore, from sin(A - B) = \(\frac{1}{2}\), we conclude:

A - B = \(\frac{\pi}{6}\) (Equation 2)

Step 3: Solve the System of Equations for A and B

We now have a system of two linear equations with A and B:

• A + B = \(\frac{\pi}{2}\) (Equation 1)
• A - B = \(\frac{\pi}{6}\) (Equation 2)

We can solve this system by adding the two equations:

(A + B) + (A - B) = \(\frac{\pi}{2} + \frac{\pi}{6}\)

2A = \(\frac{3\pi}{6} + \frac{\pi}{6}\)

2A = \(\frac{4\pi}{6}\)

2A = \(\frac{2\pi}{3}\)

A = \(\frac{\pi}{3}\)

Now substitute the value of A into Equation 1 to find B:

\(\frac{\pi}{3}\) + B = \(\frac{\pi}{2}\)

B = \(\frac{\pi}{2} - \frac{\pi}{3}\)

B = \(\frac{3\pi}{6} - \frac{2\pi}{6}\)

B = \(\frac{\pi}{6}\)

Let's check if these values satisfy the original constraints: 0 < \(\frac{\pi}{3}\) < \(\frac{\pi}{2}\) and 0 < \(\frac{\pi}{6}\) < \(\frac{\pi}{2}\). Both conditions are met (\(\frac{\pi}{3} \approx 60^\circ\), \(\frac{\pi}{6} \approx 30^\circ\), \(\frac{\pi}{2} \approx 90^\circ\)).

Step 4: Calculate tan A and tan B

Now that we have the values for A and B, we can calculate their tangents.

tan A = tan(\(\frac{\pi}{3}\))

tan A = \(\sqrt{3}\)

tan B = tan(\(\frac{\pi}{6}\))

tan B = \(\frac{1}{\sqrt{3}}\)

Step 5: Find the Ratio tan A : tan B

Finally, we need to find the ratio \(\frac{\text{tan A}}{\text{tan B}}\).

\(\frac{\text{tan A}}{\text{tan B}} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}\)

\(\frac{\text{tan A}}{\text{tan B}} = \sqrt{3} \times \sqrt{3}\)

\(\frac{\text{tan A}}{\text{tan B}} = 3\)

The ratio tan A : tan B is 3 : 1.

Summary of Results

From the given conditions, we found the specific values of angles A and B, and then calculated their tangents to find the required ratio.

The ratio tan A : tan B is \(\sqrt{3}\) : \(\frac{1}{\sqrt{3}}\), which simplifies to 3 : 1.

Revision Table: Key Trigonometric Values

Additional Information: Solving Trigonometric Equations with Constraints

When solving trigonometric equations like sin(x) = k or cos(x) = k, it's crucial to consider the given domain or constraints on the angle x. The general solutions involve periodic functions, but the constraints narrow down the possible values of x.

• For sin(x) = 1, the general solution is \(x = \frac{\pi}{2} + 2n\pi\). If x is restricted to (0, \(\pi\)), the only solution is \(x = \frac{\pi}{2}\).
• For sin(x) = \(\frac{1}{2}\), the general solutions are \(x = \frac{\pi}{6} + 2n\pi\) and \(x = \frac{5\pi}{6} + 2n\pi\). If x is restricted to (\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)), the only solution is \(x = \frac{\pi}{6}\). The other potential candidate, \(\frac{5\pi}{6}\) (or its equivalents), falls outside this range. Also, negative values like \(-\frac{7\pi}{6}\) etc. are also outside the range.

In this problem, the constraints 0 < A < \(\frac{\pi}{2}\) and 0 < B < \(\frac{\pi}{2}\) were essential for uniquely determining A + B and A - B, which in turn allowed us to find the specific values of A and B.