`tan^(-1)(x/y)-tan^(-1)(x-y)/(x+y)`is equal to(A) `pi/2` (B) `pi/3` (C) `pi/4` (D) `(-3pi)/4`

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(-3pi)/(4)`

Explanation

To solve the expression \( \tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right) \), we can use the identity for the difference of two inverse tangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]

where \( a = \frac{x}{y} \) and \( b = \frac{x-y}{x+y} \).

### Step 1: Identify \( a \) and \( b \)

Let:
\[ a = \frac{x}{y} \]
\[ b = \frac{x-y}{x+y} \]

### Step 2: Calculate \( a - b \)

Now, we need to calculate \( a - b \):
\[ a - b = \frac{x}{y} - \frac{x-y}{x+y} \]

To subtract these fractions, we need a common denominator:
\[ = \frac{x(x+y) - y(x-y)}{y(x+y)} \]

Now simplify the numerator:
\[ = \frac{x^2 + xy - (yx - y^2)}{y(x+y)} = \frac{x^2 + xy - yx + y^2}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)} \]

### Step 3: Calculate \( 1 + ab \)

Next, we calculate \( 1 + ab \):
\[ ab = \frac{x}{y} \cdot \frac{x-y}{x+y} = \frac{x(x-y)}{y(x+y)} = \frac{x^2 - xy}{y(x+y)} \]
Thus,
\[ 1 + ab = 1 + \frac{x^2 - xy}{y(x+y)} = \frac{y(x+y) + x^2 - xy}{y(x+y)} = \frac{yx + y^2 + x^2 - xy}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)} \]

### Step 4: Substitute into the identity

Now, we substitute \( a - b \) and \( 1 + ab \) into the identity:
\[ \tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}\left(\frac{\frac{x^2 + y^2}{y(x+y)}}{\frac{x^2 + y^2}{y(x+y)}}\right) = \tan^{-1}(1) \]

### Step 5: Final Result

Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we conclude that:
\[ \tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right) = \frac{\pi}{4} \]

Thus, the answer is \( \frac{\pi}{4} \), which corresponds to option (C).