Question

Find  \(tan^{-1}[\frac{(1-sin\,x+cos\,x)}{(1+sin\,x-cos\,x)}]\) = ?

Answer 1

 

3. Simplify the Numerator and Denominator:

Numerator:
\(\cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) + \cos^2(x/2) - \sin^2(x/2) = 2\cos^2(x/2) - 2\sin(x/2)\cos(x/2) = 2\cos(x/2)[\cos(x/2) - \sin(x/2)]\)

Denominator:
\(\cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) - \cos^2(x/2) + \sin^2(x/2) = 2\sin^2(x/2) + 2\sin(x/2)\cos(x/2) = 2\sin(x/2)[\sin(x/2) + \cos(x/2)]\)

4. Substitute back into the expression:

Now substitute the simplified numerator and denominator back into the expression:

\[ \tan^{-1} \left( \frac{2\cos(x/2)[\cos(x/2) - \sin(x/2)]}{2\sin(x/2)[\sin(x/2) + \cos(x/2)]} \right) \] Simplifying further: \[ \tan^{-1} \left( \frac{\cos(x/2)[\cos(x/2) - \sin(x/2)]}{\sin(x/2)[\sin(x/2) + \cos(x/2)]} \right) \]

 

5. Divide the numerator and denominator by \( \cos(x/2) \):

We can divide both the numerator and denominator by \( \cos(x/2) \), yielding:

\[ \tan^{-1} \left( \frac{\cos(x/2) - \sin(x/2)}{\sin(x/2) + \cos(x/2)} \right) \] Now, recall that: \[ 1 - \tan(x/2) = \cos(x/2) - \sin(x/2), \quad 1 + \tan(x/2) = \sin(x/2) + \cos(x/2) \] So, we can express the above equation as: \[ \tan^{-1} \left( \frac{1 - \tan(x/2)}{1 + \tan(x/2)} \right) \]

 

6. Recognize the tangent difference formula:

We know that: \[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \] Therefore, we have: \[ \tan^{-1} \left( \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \right) \]

7. Simplify the Inverse Tangent:

Since \( \tan^{-1}(\tan(\theta)) = \theta \), we have: \[ \frac{\pi}{4} - \frac{x}{2} \]

Final Answer:

Therefore, the simplified expression is: \[ \tan^{-1} \left( \frac{1 - \sin(x) + \cos(x)}{1 + \sin(x) - \cos(x)} \right) = \frac{\pi}{4} - \frac{x}{2} \]