Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
