Answer :
Sure, let's solve the given trigonometric equation step-by-step. We want to verify if the left-hand side (LHS) is equal to the right-hand side (RHS) of the equation:
[tex]\[ \frac{1 - \tan(A)}{1 + \tan(A)} = \frac{\cot(A) - 1}{\cot(A) + 1} \][/tex]
### Step 1: Express everything in terms of sine and cosine
First, recall the definitions of tangent and cotangent:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \quad \text{and} \quad \cot(A) = \frac{\cos(A)}{\sin(A)} \][/tex]
### Step 2: Substitute definitions into the LHS
Substitute [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex] in the LHS:
[tex]\[ \frac{1 - \frac{\sin(A)}{\cos(A)}}{1 + \frac{\sin(A)}{\cos(A)}} = \frac{\frac{\cos(A) - \sin(A)}{\cos(A)}}{\frac{\cos(A) + \sin(A)}{\cos(A)}} = \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 3: Substitute definitions into the RHS
Substitute [tex]\(\cot(A) = \frac{\cos(A)}{\sin(A)}\)[/tex] in the RHS:
[tex]\[ \frac{\frac{\cos(A)}{\sin(A)} - 1}{\frac{\cos(A)}{\sin(A)} + 1} = \frac{\frac{\cos(A) - \sin(A)}{\sin(A)}}{\frac{\cos(A) + \sin(A)}{\sin(A)}} = \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 4: Simplify both sides
Using the results from the substitutions:
LHS simplifies to:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
RHS simplifies to:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 5: Conclusion
Both sides simplify to the same expression:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
However, the initial and simplified expressions are indeed mathematically equivalent as shown in the process of conversion to sine and cosine terms. Checking better into the nature of the equation, we see that both, indeed, are not algebraically consistent across all domains. So we have them identically simplified but existing necessity to indicate they are fundamentally consistent:
[tex]\[ \text{The original equation does not hold consistently true across all values of } A. \][/tex]
Thus, [tex]\(\frac{1 - \tan(A)}{1 + \tan(A)} \neq \frac{\cot(A) - 1}{\cot(A) + 1}\)[/tex] generically as the logical closure. The simplified terms seem equivalent in stepwork because both conversions match but as per contextural structures, globally can't valid.
So the complicated step confirms:
The given trigonometric identity doesn't hold true consistently.
[tex]\[ \frac{1 - \tan(A)}{1 + \tan(A)} = \frac{\cot(A) - 1}{\cot(A) + 1} \][/tex]
### Step 1: Express everything in terms of sine and cosine
First, recall the definitions of tangent and cotangent:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \quad \text{and} \quad \cot(A) = \frac{\cos(A)}{\sin(A)} \][/tex]
### Step 2: Substitute definitions into the LHS
Substitute [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex] in the LHS:
[tex]\[ \frac{1 - \frac{\sin(A)}{\cos(A)}}{1 + \frac{\sin(A)}{\cos(A)}} = \frac{\frac{\cos(A) - \sin(A)}{\cos(A)}}{\frac{\cos(A) + \sin(A)}{\cos(A)}} = \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 3: Substitute definitions into the RHS
Substitute [tex]\(\cot(A) = \frac{\cos(A)}{\sin(A)}\)[/tex] in the RHS:
[tex]\[ \frac{\frac{\cos(A)}{\sin(A)} - 1}{\frac{\cos(A)}{\sin(A)} + 1} = \frac{\frac{\cos(A) - \sin(A)}{\sin(A)}}{\frac{\cos(A) + \sin(A)}{\sin(A)}} = \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 4: Simplify both sides
Using the results from the substitutions:
LHS simplifies to:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
RHS simplifies to:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
### Step 5: Conclusion
Both sides simplify to the same expression:
[tex]\[ \frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)} \][/tex]
However, the initial and simplified expressions are indeed mathematically equivalent as shown in the process of conversion to sine and cosine terms. Checking better into the nature of the equation, we see that both, indeed, are not algebraically consistent across all domains. So we have them identically simplified but existing necessity to indicate they are fundamentally consistent:
[tex]\[ \text{The original equation does not hold consistently true across all values of } A. \][/tex]
Thus, [tex]\(\frac{1 - \tan(A)}{1 + \tan(A)} \neq \frac{\cot(A) - 1}{\cot(A) + 1}\)[/tex] generically as the logical closure. The simplified terms seem equivalent in stepwork because both conversions match but as per contextural structures, globally can't valid.
So the complicated step confirms:
The given trigonometric identity doesn't hold true consistently.
