Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Alright, let's analyze each of the given expressions to determine which of them are trigonometric identities.
### Expression A:
\(\tan^2 x = 1 + \sec^2 x\)
To evaluate this expression, let's use some known trigonometric identities:
[tex]\[ \tan^2 x = \sec^2 x - 1 \][/tex]
Plugging this into the left-hand side:
[tex]\[ \tan^2 x = \sec^2 x - 1 \][/tex]
Thus:
[tex]\[ \sec^2 x - 1 = 1 + \sec^2 x \][/tex]
Here, subtracting \(\sec^2 x\) from both sides yields:
[tex]\[ -1 = 1 \][/tex]
This is a contradiction, so \( \tan^2 x \neq 1 + \sec^2 x \).
Therefore, Expression A is not an identity.
### Expression B:
\(\sin^2 x = 1 - \cos^2 x\)
Rewriting \(1 - \cos^2 x\):
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
This follows directly from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Subtracting \(\cos^2 x\) from both sides gives:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Therefore, Expression B is an identity.
### Expression C:
\(\sin^2 x - \cos^2 x = 1\)
Starting from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Let's test the expression by substituting known angles. For example, let \(x = 0\):
[tex]\[ \sin^2 (0) - \cos^2 (0) = 0 - 1 = -1 \neq 1 \][/tex]
Therefore, the given equation does not hold for \(x = 0\).
Thus, Expression C is not an identity.
### Expression D:
\(\cot^2 x = \csc^2 x - 1\)
To analyze this, let's use known identities:
[tex]\[ \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \][/tex]
and
[tex]\[ \csc^2 x = \frac{1}{\sin^2 x} \][/tex]
Rewriting the right-hand side:
[tex]\[ \csc^2 x - 1 = \frac{1}{\sin^2 x} - 1 \][/tex]
To combine into a single fraction:
[tex]\[ \frac{1}{\sin^2 x} - 1 = \frac{1 - \sin^2 x}{\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \][/tex]
Hence:
[tex]\[ \cot^2 x = \csc^2 x - 1 \][/tex]
Therefore, Expression D is an identity.
### Conclusion
Based on our evaluations, the correct identities are:
- B. \(\sin^2 x = 1 - \cos^2 x\)
- D. \(\cot^2 x = \csc^2 x - 1\)
The identities are B and D.
### Expression A:
\(\tan^2 x = 1 + \sec^2 x\)
To evaluate this expression, let's use some known trigonometric identities:
[tex]\[ \tan^2 x = \sec^2 x - 1 \][/tex]
Plugging this into the left-hand side:
[tex]\[ \tan^2 x = \sec^2 x - 1 \][/tex]
Thus:
[tex]\[ \sec^2 x - 1 = 1 + \sec^2 x \][/tex]
Here, subtracting \(\sec^2 x\) from both sides yields:
[tex]\[ -1 = 1 \][/tex]
This is a contradiction, so \( \tan^2 x \neq 1 + \sec^2 x \).
Therefore, Expression A is not an identity.
### Expression B:
\(\sin^2 x = 1 - \cos^2 x\)
Rewriting \(1 - \cos^2 x\):
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
This follows directly from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Subtracting \(\cos^2 x\) from both sides gives:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Therefore, Expression B is an identity.
### Expression C:
\(\sin^2 x - \cos^2 x = 1\)
Starting from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Let's test the expression by substituting known angles. For example, let \(x = 0\):
[tex]\[ \sin^2 (0) - \cos^2 (0) = 0 - 1 = -1 \neq 1 \][/tex]
Therefore, the given equation does not hold for \(x = 0\).
Thus, Expression C is not an identity.
### Expression D:
\(\cot^2 x = \csc^2 x - 1\)
To analyze this, let's use known identities:
[tex]\[ \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \][/tex]
and
[tex]\[ \csc^2 x = \frac{1}{\sin^2 x} \][/tex]
Rewriting the right-hand side:
[tex]\[ \csc^2 x - 1 = \frac{1}{\sin^2 x} - 1 \][/tex]
To combine into a single fraction:
[tex]\[ \frac{1}{\sin^2 x} - 1 = \frac{1 - \sin^2 x}{\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \][/tex]
Hence:
[tex]\[ \cot^2 x = \csc^2 x - 1 \][/tex]
Therefore, Expression D is an identity.
### Conclusion
Based on our evaluations, the correct identities are:
- B. \(\sin^2 x = 1 - \cos^2 x\)
- D. \(\cot^2 x = \csc^2 x - 1\)
The identities are B and D.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.