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: Observe what is required.
We are given \[ \tan(x-y)=\frac{4}{5} \quad \text{and} \quad \tan(x+y)=\frac{6}{5} \] We need to find \[ \tan 2x \] Now notice that \[ 2x=(x-y)+(x+y) \] So we can use the tangent addition formula.

Step 2: Use the tangent addition identity.
We know that \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \] Take \[ A=x-y,\qquad B=x+y \] Then \[ A+B=(x-y)+(x+y)=2x \] Therefore, \[ \tan 2x=\frac{\tan(x-y)+\tan(x+y)}{1-\tan(x-y)\tan(x+y)} \]

Step 3: Substitute the given values.
Substituting \[ \tan(x-y)=\frac{4}{5},\qquad \tan(x+y)=\frac{6}{5} \] we get \[ \tan 2x=\frac{\frac{4}{5}+\frac{6}{5}}{1-\left(\frac{4}{5}\cdot\frac{6}{5}\right)} \] \[ =\frac{\frac{10}{5}}{1-\frac{24}{25}} \] \[ =\frac{2}{\frac{1}{25}} \]

Step 4: Simplify the expression.
Now divide: \[ \frac{2}{\frac{1}{25}}=2 \times 25=50 \] Hence, \[ \tan 2x=50 \]

Step 5: Check the denominator carefully.
Let us verify: \[ 1-\frac{24}{25}=\frac{25-24}{25}=\frac{1}{25} \] So the denominator is indeed positive and non-zero, hence the calculation is valid.

Step 6: Use the condition \( 0<x,y<\dfrac{\pi}{4} \).
The condition ensures that the angles involved remain in ranges where the tangent values are well-defined and positive as given.
So there is no ambiguity in sign while using the formula.

Step 7: Final conclusion.
Therefore, \[ \boxed{\tan 2x=50} \] Hence, the correct option is \[ \boxed{(4)\ 50} \]