Certainly! Let's carefully go through the given equation and demonstrate that both sides are indeed equal, establishing it as an identity.
We need to show that:
[tex]\[
\frac{\cot \theta - 1}{\cot \theta - 1} = \frac{1 - \tan \theta}{1 + \tan \theta}
\][/tex]
### Step-by-Step Solution:
1. Examine the left-hand side (LHS):
[tex]\[
\frac{\cot \theta - 1}{\cot \theta - 1}
\][/tex]
Here, provided that [tex]\(\cot \theta \neq 1\)[/tex], [tex]\(\cot \theta - 1\)[/tex] will not be zero, and thus:
[tex]\[
\frac{\cot \theta - 1}{\cot \theta - 1} = 1
\][/tex]
This is because any non-zero quantity divided by itself is equal to 1.
2. Examine the right-hand side (RHS):
[tex]\[
\frac{1 - \tan \theta}{1 + \tan \theta}
\][/tex]
We need to show that this also simplifies to 1. Let's simplify it further.
3. Simplify RHS by substituting [tex]\(\tan \theta\)[/tex]:
Recall that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex]. By substituting [tex]\(\tan \theta = \frac{1}{\cot \theta}\)[/tex]:
[tex]\[
\frac{1 - \frac{1}{\cot \theta}}{1 + \frac{1}{\cot \theta}}
\][/tex]
4. Combine fractions in the numerator and the denominator:
Numerator:
[tex]\[
1 - \frac{1}{\cot \theta} = \frac{\cot \theta - 1}{\cot \theta}
\][/tex]
Denominator:
[tex]\[
1 + \frac{1}{\cot \theta} = \frac{\cot \theta + 1}{\cot \theta}
\][/tex]
5. Rewrite the RHS combining the two fractions:
[tex]\[
\frac{\frac{\cot \theta - 1}{\cot \theta}}{\frac{\cot \theta + 1}{\cot \theta}} = \frac{\cot \theta - 1}{\cot \theta + 1}
\][/tex]
### Conclusion:
- From earlier, we found that the LHS simplifies to 1.
- From the simplification of the RHS, note that it simplifies to 1 too, provided that [tex]\(\cot \theta\)[/tex] is not equal to 1 and not equal to -1.
Thus, we verify the equation:
[tex]\[
\frac{\cot \theta - 1}{\cot \theta - 1} = \frac{1 - \tan \theta}{1 + \tan \theta}
\][/tex]
is indeed an identity since both sides simplify to 1.
