To verify the given trigonometric identity [tex]\(\frac{1+\tan \theta}{1-\tan \theta} = \frac{\cot \theta+1}{\cot \theta-1}\)[/tex], let's proceed step-by-step as follows:
1. Left-hand side (LHS):
[tex]\[
\text{LHS} = \frac{1 + \tan \theta}{1 - \tan \theta}
\][/tex]
2. Right-hand side (RHS):
[tex]\[
\text{RHS} = \frac{\cot \theta + 1}{\cot \theta - 1}
\][/tex]
3. Express [tex]\(\cot \theta\)[/tex] in terms of [tex]\(\tan \theta\)[/tex]:
Recall that:
[tex]\[
\cot \theta = \frac{1}{\tan \theta}
\][/tex]
4. Substitute [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex] into the RHS:
[tex]\[
\text{RHS} = \frac{\frac{1}{\tan \theta} + 1}{\frac{1}{\tan \theta} - 1}
\][/tex]
5. Simplify the expression on the RHS:
[tex]\[
\text{RHS} = \frac{\frac{1 + \tan \theta}{\tan \theta}}{\frac{1 - \tan \theta}{\tan \theta}}
\][/tex]
Cancel the common factor [tex]\(\frac{1}{\tan \theta}\)[/tex] in the numerator and the denominator:
[tex]\[
\text{RHS} = \frac{1 + \tan \theta}{1 - \tan \theta}
\][/tex]
6. Compare LHS with RHS:
[tex]\[
\text{LHS} = \frac{1 + \tan \theta}{1 - \tan \theta} = \text{RHS}
\][/tex]
We have shown that both sides are indeed equal. Thus, the given identity [tex]\(\frac{1 + \tan \theta}{1 - \tan \theta} = \frac{\cot \theta + 1}{\cot \theta - 1}\)[/tex] is true.
Therefore, the trigonometric identity holds true:
[tex]\[
\frac{1 + \tan \theta}{1 - \tan \theta} = \frac{\cot \theta + 1}{\cot \theta - 1}.
\][/tex]
