If `0 lt A , B lt 45^(@)` cos (A+B) = 24 /25 and sin (A-B) = 15/17 then tan 2 A is

Options

A. A.0
B. B.1
C. C.`213//4`
D. D.`416//87`

Answer

To solve the problem step by step, we need to find the value of \( \tan 2A \) given the conditions:

1. \( \cos(A + B) = \frac{24}{25} \)
2. \( \sin(A - B) = \frac{15}{17} \)

### Step 1: Find \( \sin(A + B) \) and \( \sin(A - B) \)

Using the Pythagorean identity, we can find \( \sin(A + B) \) and \( \cos(A + B) \):

\[ \sin^2(A + B) + \cos^2(A + B) = 1 \]

Given \( \cos(A + B) = \frac{24}{25} \):

\[ \sin^2(A + B) = 1 - \left(\frac{24}{25}\right)^2 = 1 - \frac{576}{625} = \frac{49}{625} \]

Thus,

\[ \sin(A + B) = \frac{7}{25} \]

Similarly, for \( \sin(A - B) \):

Given \( \sin(A - B) = \frac{15}{17} \):

Using the Pythagorean identity again:

\[ \cos^2(A - B) = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289} \]

Thus,

\[ \cos(A - B) = \frac{8}{17} \]

### Step 2: Use the tangent addition formula

We can find \( \tan A \) and \( \tan B \) using the following formulas:

\[ \tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\frac{7}{25}}{\frac{24}{25}} = \frac{7}{24} \]

\[ \tan(A - B) = \frac{\sin(A - B)}{\cos(A - B)} = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{8} \]

### Step 3: Use the formula for \( \tan 2A \)

Using the formula for \( \tan 2A \):

\[ \tan 2A = \frac{\tan(A + B) + \tan(A - B)}{1 - \tan(A + B) \tan(A - B)} \]

Substituting the values we found:

\[ \tan 2A = \frac{\frac{7}{24} + \frac{15}{8}}{1 - \left(\frac{7}{24} \cdot \frac{15}{8}\right)} \]

### Step 4: Simplify the expression

First, find a common denominator for the numerator:

\[ \frac{7}{24} + \frac{15}{8} = \frac{7}{24} + \frac{45}{24} = \frac{52}{24} = \frac{13}{6} \]

Now calculate the denominator:

\[ 1 - \left(\frac{7}{24} \cdot \frac{15}{8}\right) = 1 - \frac{105}{192} = \frac{192 - 105}{192} = \frac{87}{192} \]

### Step 5: Final calculation

Now substituting back into the formula for \( \tan 2A \):

\[ \tan 2A = \frac{\frac{13}{6}}{\frac{87}{192}} = \frac{13 \cdot 192}{6 \cdot 87} = \frac{13 \cdot 32}{87} = \frac{416}{87} \]

Thus, the final answer is:

\[ \tan 2A = \frac{416}{87} \]