Prove that :
`(tan theta)/(1 - cot theta) + (cot theta)/(1 - tan theta) = 1 + sec theta cosec theta` .

Answer

To prove the identity

\[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta, \]

we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Rewrite \(\tan \theta\) and \(\cot \theta\)

Recall the definitions of tangent and cotangent:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}. \]

Substituting these definitions into the LHS gives:

\[ \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}. \]

### Step 2: Simplify the denominators

For the first term:

\[ 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta}. \]

Thus, the first term becomes:

\[ \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta - \cos \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}. \]

For the second term:

\[ 1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta}. \]

Thus, the second term becomes:

\[ \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta - \sin \theta} = \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}. \]

### Step 3: Combine the two fractions

Now we have:

\[ \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}. \]

Notice that \(\cos \theta - \sin \theta = -(\sin \theta - \cos \theta)\). Thus, we can rewrite the second term:

\[ \frac{\cos^2 \theta}{\sin \theta (-1)(\sin \theta - \cos \theta)} = -\frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}. \]

Now we can combine the fractions:

\[ \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}. \]

### Step 4: Use the identity \( \sin^2 \theta - \cos^2 \theta \)

Using the identity \( \sin^2 \theta - \cos^2 \theta = -(\cos^2 \theta - \sin^2 \theta) \), we can rewrite this as:

\[ \frac{-(\cos^2 \theta - \sin^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}. \]

### Step 5: Factor the numerator

The numerator can be factored as:

\[ -(\sin \theta - \cos \theta)(\sin \theta + \cos \theta). \]

Thus, we have:

\[ \frac{-(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}. \]

The \((\sin \theta - \cos \theta)\) terms cancel out:

\[ \frac{-(\sin \theta + \cos \theta)}{\sin \theta \cos \theta}. \]

### Step 6: Rewrite in terms of secant and cosecant

Now, we can express this as:

\[ -\left(\frac{1}{\sin \theta} + \frac{1}{\cos \theta}\right) = -\left(\csc \theta + \sec \theta\right). \]

Finally, we rewrite the LHS as:

\[ 1 + \sec \theta \csc \theta. \]

### Conclusion

Thus, we have shown that:

\[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta. \]

Hence, the identity is proved.