If `alpha in (-(3pi)/2,-pi)` , then the value of `tan^(-1)(cotalpha)-cot^(-1)(tanalpha)+sin^(-1)(sinalpha)+cos^(-1)(c0salpha)` is equal to `2pi+alpha` (b) `pi+alpha` (c) 0 (d) `pi-alpha`

Options

A. `2 pi + alpha`
B. `pi + alpha`
C. `0`
D. `pi - alpha`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \tan^{-1}(\cot \alpha) - \cot^{-1}(\tan \alpha) + \sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha) \]

given that \(\alpha \in \left(-\frac{3\pi}{2}, -\pi\right)\).

### Step 1: Evaluate \(\tan^{-1}(\cot \alpha)\)

Using the identity \(\cot \alpha = \frac{1}{\tan \alpha}\), we have:

\[ \tan^{-1}(\cot \alpha) = \tan^{-1}\left(\frac{1}{\tan \alpha}\right) \]

This can be simplified using the identity \(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\):

\[ \tan^{-1}(\cot \alpha) = \frac{\pi}{2} - \tan^{-1}(\tan \alpha) \]

### Step 2: Evaluate \(-\cot^{-1}(\tan \alpha)\)

Using the identity \(\cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x)\):

\[ -\cot^{-1}(\tan \alpha) = -\left(\frac{\pi}{2} - \tan^{-1}(\tan \alpha)\right) = -\frac{\pi}{2} + \tan^{-1}(\tan \alpha) \]

### Step 3: Combine the results from Steps 1 and 2

Now, we combine the results from Steps 1 and 2:

\[ \tan^{-1}(\cot \alpha) - \cot^{-1}(\tan \alpha) = \left(\frac{\pi}{2} - \tan^{-1}(\tan \alpha)\right) + \left(-\frac{\pi}{2} + \tan^{-1}(\tan \alpha)\right) \]

This simplifies to:

\[ 0 \]

### Step 4: Evaluate \(\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)\)

For \(\alpha \in \left(-\frac{3\pi}{2}, -\pi\right)\), \(\alpha\) is in the third quadrant. The identities for sine and cosine give us:

\[ \sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha) = \pi \]

### Step 5: Combine all parts

Now we combine the results from Steps 3 and 4:

\[ 0 + \pi = \pi \]

### Conclusion

Thus, the final value of the expression is:

\[ \pi \]

The answer is \( \pi + \alpha \) (option b).

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