Show that `2tan^-1 (-3)=-pi/2+tan^-1(-4/3)`

Explanation

To prove that \( 2 \tan^{-1}(-3) = -\frac{\pi}{2} + \tan^{-1}\left(-\frac{4}{3}\right) \), we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Start with the LHS
\[ LHS = 2 \tan^{-1}(-3) \]

### Step 2: Use the property of the inverse tangent function
We know that:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
Applying this property, we have:
\[ LHS = 2 \tan^{-1}(-3) = 2 \cdot (-\tan^{-1}(3)) = -2 \tan^{-1}(3) \]

### Step 3: Use the double angle formula for tangent
The double angle formula for tangent states:
\[ 2 \tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
Using this formula, we can rewrite \(-2 \tan^{-1}(3)\):
\[ -2 \tan^{-1}(3) = -\tan^{-1}\left(\frac{2 \cdot 3}{1 - 3^2}\right) = -\tan^{-1}\left(\frac{6}{1 - 9}\right) = -\tan^{-1}\left(\frac{6}{-8}\right) = -\tan^{-1}\left(-\frac{3}{4}\right) \]

### Step 4: Use the property of the inverse tangent function again
Using the property again:
\[ -\tan^{-1}\left(-\frac{3}{4}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]
Thus, we have:
\[ LHS = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 5: Now, let's evaluate the RHS
\[ RHS = -\frac{\pi}{2} + \tan^{-1}\left(-\frac{4}{3}\right) \]
Using the property of the inverse tangent function:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
We can rewrite the RHS:
\[ RHS = -\frac{\pi}{2} - \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Use the identity for addition of angles
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
If we let \( a = \frac{3}{4} \) and \( b = \frac{4}{3} \):
\[ ab = \frac{3}{4} \cdot \frac{4}{3} = 1 \]
Thus:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{4}{3}\right) = \frac{\pi}{2} \]
So,
\[ \tan^{-1}\left(\frac{3}{4}\right) = \frac{\pi}{2} - \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 7: Substitute back into the RHS
Now we can substitute this back into the RHS:
\[ RHS = -\frac{\pi}{2} - \left(\frac{\pi}{2} - \tan^{-1}\left(\frac{4}{3}\right)\right) = -\frac{\pi}{2} - \frac{\pi}{2} + \tan^{-1}\left(\frac{4}{3}\right) = -\pi + \tan^{-1}\left(\frac{4}{3}\right) \]

### Conclusion
Since both sides are equal:
\[ LHS = RHS \]
Thus, we have shown that:
\[ 2 \tan^{-1}(-3) = -\frac{\pi}{2} + \tan^{-1}\left(-\frac{4}{3}\right) \]