Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Certainly! To verify the identity [tex]\(\tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta)\)[/tex], let's carefully go through the expressions step by step.
1. Left-hand side (LHS):
[tex]\[ \tan^2 \theta - \cot^2 \theta \][/tex]
We know that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex], thus:
[tex]\[ \cot^2 \theta = \left( \frac{1}{\tan \theta} \right)^2 = \frac{1}{\tan^2 \theta} \][/tex]
Therefore, substituting [tex]\(\cot^2 \theta\)[/tex] in the LHS:
[tex]\[ \tan^2 \theta - \frac{1}{\tan^2 \theta} \][/tex]
2. Right-hand side (RHS):
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
We know that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex] and using [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Therefore, substituting [tex]\(\sec^2 \theta\)[/tex] and simplifying [tex]\(1 - \cot^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta \left(1 - \frac{1}{\tan^2 \theta}\right) \][/tex]
3. Simplifying inside the parentheses on the RHS:
[tex]\[ 1 - \frac{1}{\tan^2 \theta} = \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
Thus, the RHS expression becomes:
[tex]\[ \sec^2 \theta \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
We know:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Substituting [tex]\(\sec^2 \theta\)[/tex] back in:
[tex]\[ (1 + \tan^2 \theta) \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
Rewriting the RHS fully:
[tex]\[ (1 + \tan^2 \theta) \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
We get the product:
[tex]\[ \frac{(\tan^2 \theta - 1)(1 + \tan^2 \theta)}{\tan^2 \theta} \][/tex]
After performing these steps and comparing the simplified forms, we see that:
- The LHS simplifies to [tex]\(\tan^2 \theta - \frac{1}{\tan^2 \theta}\)[/tex]
- The RHS simplifies to [tex]\(\frac{(\tan^2 \theta - 1)(1 + \tan^2 \theta)}{\tan^2 \theta}\)[/tex]
Upon cross-comparison, it is clear that [tex]\(\tan^2 \theta - \cot^2 \theta\)[/tex] and [tex]\(\sec^2 \theta (1 - \cot^2 \theta)\)[/tex] are not identical expressions for the given function properties. Therefore, the original equation:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
is not valid for all [tex]\(\theta\)[/tex].
Hence, we conclude:
The statement [tex]\(\tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta)\)[/tex] is False.
1. Left-hand side (LHS):
[tex]\[ \tan^2 \theta - \cot^2 \theta \][/tex]
We know that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex], thus:
[tex]\[ \cot^2 \theta = \left( \frac{1}{\tan \theta} \right)^2 = \frac{1}{\tan^2 \theta} \][/tex]
Therefore, substituting [tex]\(\cot^2 \theta\)[/tex] in the LHS:
[tex]\[ \tan^2 \theta - \frac{1}{\tan^2 \theta} \][/tex]
2. Right-hand side (RHS):
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
We know that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex] and using [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Therefore, substituting [tex]\(\sec^2 \theta\)[/tex] and simplifying [tex]\(1 - \cot^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta \left(1 - \frac{1}{\tan^2 \theta}\right) \][/tex]
3. Simplifying inside the parentheses on the RHS:
[tex]\[ 1 - \frac{1}{\tan^2 \theta} = \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
Thus, the RHS expression becomes:
[tex]\[ \sec^2 \theta \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
We know:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Substituting [tex]\(\sec^2 \theta\)[/tex] back in:
[tex]\[ (1 + \tan^2 \theta) \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
Rewriting the RHS fully:
[tex]\[ (1 + \tan^2 \theta) \cdot \frac{\tan^2 \theta - 1}{\tan^2 \theta} \][/tex]
We get the product:
[tex]\[ \frac{(\tan^2 \theta - 1)(1 + \tan^2 \theta)}{\tan^2 \theta} \][/tex]
After performing these steps and comparing the simplified forms, we see that:
- The LHS simplifies to [tex]\(\tan^2 \theta - \frac{1}{\tan^2 \theta}\)[/tex]
- The RHS simplifies to [tex]\(\frac{(\tan^2 \theta - 1)(1 + \tan^2 \theta)}{\tan^2 \theta}\)[/tex]
Upon cross-comparison, it is clear that [tex]\(\tan^2 \theta - \cot^2 \theta\)[/tex] and [tex]\(\sec^2 \theta (1 - \cot^2 \theta)\)[/tex] are not identical expressions for the given function properties. Therefore, the original equation:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
is not valid for all [tex]\(\theta\)[/tex].
Hence, we conclude:
The statement [tex]\(\tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta)\)[/tex] is False.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.