Question

If $\tan(x-y)=\frac{4}{5}$, $\tan(x+y)=\frac{6}{5}$ and $0<x,y<\frac{\pi}{4}$, then $\tan 2x$ is:

• A. \( 62 \)
• B. \( 60 \)
• C. \( 54 \)
• D. \( 50 \)
• E. \( 55 \)

Hint:

Whenever expressions like \( x-y \) and \( x+y \) are given together, first look for their sum or difference. It often leads directly to identities like \( 2x \) or \( 2y \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We are given the values of \(\tan(A)\) and \(\tan(B)\) where \(A=x-y\) and \(B=x+y\). We need to find \(\tan(2x)\). We can express \(2x\) as a sum of \(A\) and \(B\), i.e., \(2x = (x-y) + (x+y)\).
Step 2: Key Formula or Approach:
We will use the tangent addition formula:
\[ \tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \] Here, \(A = x-y\) and \(B = x+y\). So, we want to find \(\tan((x-y)+(x+y)) = \tan(2x)\).
Step 3: Detailed Explanation:
Let \(A = x-y\) and \(B = x+y\). We are given \(\tan(A) = \frac{4}{5}\) and \(\tan(B) = \frac{6}{5}\).
Using the formula for \(\tan(A+B)\):
\[ \tan(2x) = \tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \] Substitute the given values:
\[ \tan(2x) = \frac{\frac{4}{5} + \frac{6}{5}}{1 - (\frac{4}{5})(\frac{6}{5})} \] Simplify the numerator:
\[ \frac{4}{5} + \frac{6}{5} = \frac{10}{5} = 2 \] Simplify the denominator:
\[ 1 - \frac{24}{25} = \frac{25}{25} - \frac{24}{25} = \frac{1}{25} \] Now, calculate the final value:
\[ \tan(2x) = \frac{2}{\frac{1}{25}} = 2 \times 25 = 50 \] The condition \(0<x, y<\frac{\pi}{4}\) ensures that \(x-y\) and \(x+y\) are in ranges where the tangent function is well-behaved. \(0<x+y<\pi/2\) and \(-\pi/4<x-y<\pi/4\). Step 4: Final Answer:
The value of \(\tan(2x)\) is 50. This matches option (D).