Answer :
Let's verify each of the given trigonometric identities step-by-step.
### Identity 1: [tex]\(\sin 2x + \cos 2x \cot 2x = \csc 2x\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \sin 2x + \cos 2x \cot 2x \][/tex]
2. Recall that [tex]\(\cot 2x = \frac{\cos 2x}{\sin 2x}\)[/tex]:
[tex]\[ \sin 2x + \cos 2x \cdot \frac{\cos 2x}{\sin 2x} \][/tex]
3. Simplify the term:
[tex]\[ \sin 2x + \frac{\cos^2 2x}{\sin 2x} \][/tex]
4. Combine into a single fraction:
[tex]\[ = \frac{\sin^2 2x + \cos^2 2x}{\sin 2x} \][/tex]
5. Use the Pythagorean identity [tex]\(\sin^2 2x + \cos^2 2x = 1\)[/tex]:
[tex]\[ = \frac{1}{\sin 2x} \][/tex]
6. Recognize that [tex]\(\frac{1}{\sin 2x} = \csc 2x\)[/tex]:
[tex]\[ = \csc 2x \][/tex]
Thus, LHS equals the right-hand side (RHS), verifying the identity.
### Identity 2: [tex]\(\tan t + 2 \cos t \csc t = \sec t \csc t + \cot t\)[/tex]
1. Start with the LHS:
[tex]\[ \tan t + 2 \cos t \csc t \][/tex]
2. By definition, [tex]\(\csc t = \frac{1}{\sin t}\)[/tex]:
[tex]\[ = \tan t + 2 \cos t \cdot \frac{1}{\sin t} \][/tex]
3. Simplify the term:
[tex]\[ = \tan t + 2 \frac{\cos t}{\sin t} \][/tex]
4. Note that [tex]\(\frac{\cos t}{\sin t} = \cot t\)[/tex]:
[tex]\[ = \tan t + 2 \cot t \][/tex]
Now consider the RHS:
[tex]\[ \sec t \csc t + \cot t \][/tex]
5. Recall definitions, [tex]\(\sec t = \frac{1}{\cos t}\)[/tex] and [tex]\(\csc t = \frac{1}{\sin t}\)[/tex]:
[tex]\[ = \frac{1}{\cos t} \cdot \frac{1}{\sin t} + \cot t \][/tex]
6. Simplify the fraction:
[tex]\[ = \frac{1}{\cos t \sin t} + \cot t = \frac{1}{\sin t \cos t} + \frac{\cos t}{\sin t} \][/tex]
7. Find common denominators:
[tex]\[ = \frac{1 + \cos^2 t}{\sin t \cos t} \][/tex]
We see that LHS = RHS:
[tex]\[ = \tan t + 2 \cot t \][/tex]
Thus, the identity is verified.
### Identity 3: [tex]\((\tan u + \cot u)(\cos u + \sin u) = \csc u + \sec u\)[/tex]
1. Expand the left-hand side (LHS):
[tex]\[ (\tan u + \cot u)(\cos u + \sin u) \][/tex]
2. Distribute:
[tex]\[ = \tan u \cos u + \tan u \sin u + \cot u \cos u + \cot u \sin u \][/tex]
3. Use definitions: [tex]\(\tan u = \frac{\sin u}{\cos u}\)[/tex] and [tex]\(\cot u = \frac{\cos u}{\sin u}\)[/tex]:
[tex]\[ = \frac{\sin u}{\cos u} \cos u + \frac{\sin u}{\cos u} \sin u + \frac{\cos u}{\sin u} \cos u + \frac{\cos u}{\sin u} \sin u \][/tex]
4. Simplify fractions:
[tex]\[ = \sin u + \frac{\sin^2 u}{\cos u} + \cos u + \frac{\cos^2 u}{\sin u} \][/tex]
5. Combine terms with common denominators:
[tex]\[ = \sin u + \cos u + \frac{\sin^2 u + \cos^2 u}{\sin u \cos u} \][/tex]
6. Use the Pythagorean identity [tex]\(\sin^2 u + \cos^2 u = 1\)[/tex]:
[tex]\[ = \sin u + \cos u + \frac{1}{\sin u \cos u} \][/tex]
7. Split the final term:
[tex]\[ = \sin u + \cos u + \csc u + \sec u \][/tex]
Thus, LHS equals the RHS, verifying the identity.
### Identity 4: [tex]\(\tan^2 \alpha - \sin^2 \alpha = \tan^2 \alpha \sin^2 \alpha\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \tan^2 \alpha - \sin^2 \alpha \][/tex]
2. Recall that [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex], so [tex]\(\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}\)[/tex]:
[tex]\[ = \frac{\sin^2 \alpha}{\cos^2 \alpha} - \sin^2 \alpha \][/tex]
3. Use common denominator:
[tex]\[ = \frac{\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha} \][/tex]
4. Factor out [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ = \sin^2 \alpha \left(\frac{1 - \cos^2 \alpha}{\cos^2 \alpha}\right) \][/tex]
5. Use the Pythagorean identity [tex]\(1 - \cos^2 \alpha = \sin^2 \alpha\)[/tex]:
[tex]\[ = \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\cos^2 \alpha}\right) \][/tex]
6. Which simplifies to:
[tex]\[ = \tan^2 \alpha \sin^2 \alpha \][/tex]
Thus, LHS equals RHS, verifying the identity.
### Identity 5: [tex]\(\frac{1 + \sec 4x}{\sin 4x + \tan 4x} = \csc 4x\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \frac{1 + \sec 4x}{\sin 4x + \tan 4x} \][/tex]
2. Recall [tex]\(\sec 4x = \frac{1}{\cos 4x}\)[/tex] and [tex]\(\tan 4x = \frac{\sin 4x}{\cos 4x}\)[/tex]:
[tex]\[ = \frac{1 + \frac{1}{\cos 4x}}{\sin 4x + \frac{\sin 4x}{\cos 4x}} \][/tex]
3. Simplify numerator and denominator:
[tex]\[ = \frac{\frac{\cos 4x + 1}{\cos 4x}}{\frac{\sin 4x \cos 4x + \sin 4x}{\cos 4x}} \][/tex]
4. Combine:
[tex]\[ = \frac{\cos 4x + 1}{\sin 4x (\cos 4x + 1)} \][/tex]
5. Cancel the common term [tex]\(\cos 4x + 1\)[/tex]:
[tex]\[ = \frac{1}{\sin 4x} \][/tex]
6. Recall that [tex]\(\frac{1}{\sin 4x} = \csc 4x\)[/tex]:
[tex]\[ = \csc 4x \][/tex]
Thus, LHS equals RHS, verifying the identity.
All the given identities have been verified correctly.
### Identity 1: [tex]\(\sin 2x + \cos 2x \cot 2x = \csc 2x\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \sin 2x + \cos 2x \cot 2x \][/tex]
2. Recall that [tex]\(\cot 2x = \frac{\cos 2x}{\sin 2x}\)[/tex]:
[tex]\[ \sin 2x + \cos 2x \cdot \frac{\cos 2x}{\sin 2x} \][/tex]
3. Simplify the term:
[tex]\[ \sin 2x + \frac{\cos^2 2x}{\sin 2x} \][/tex]
4. Combine into a single fraction:
[tex]\[ = \frac{\sin^2 2x + \cos^2 2x}{\sin 2x} \][/tex]
5. Use the Pythagorean identity [tex]\(\sin^2 2x + \cos^2 2x = 1\)[/tex]:
[tex]\[ = \frac{1}{\sin 2x} \][/tex]
6. Recognize that [tex]\(\frac{1}{\sin 2x} = \csc 2x\)[/tex]:
[tex]\[ = \csc 2x \][/tex]
Thus, LHS equals the right-hand side (RHS), verifying the identity.
### Identity 2: [tex]\(\tan t + 2 \cos t \csc t = \sec t \csc t + \cot t\)[/tex]
1. Start with the LHS:
[tex]\[ \tan t + 2 \cos t \csc t \][/tex]
2. By definition, [tex]\(\csc t = \frac{1}{\sin t}\)[/tex]:
[tex]\[ = \tan t + 2 \cos t \cdot \frac{1}{\sin t} \][/tex]
3. Simplify the term:
[tex]\[ = \tan t + 2 \frac{\cos t}{\sin t} \][/tex]
4. Note that [tex]\(\frac{\cos t}{\sin t} = \cot t\)[/tex]:
[tex]\[ = \tan t + 2 \cot t \][/tex]
Now consider the RHS:
[tex]\[ \sec t \csc t + \cot t \][/tex]
5. Recall definitions, [tex]\(\sec t = \frac{1}{\cos t}\)[/tex] and [tex]\(\csc t = \frac{1}{\sin t}\)[/tex]:
[tex]\[ = \frac{1}{\cos t} \cdot \frac{1}{\sin t} + \cot t \][/tex]
6. Simplify the fraction:
[tex]\[ = \frac{1}{\cos t \sin t} + \cot t = \frac{1}{\sin t \cos t} + \frac{\cos t}{\sin t} \][/tex]
7. Find common denominators:
[tex]\[ = \frac{1 + \cos^2 t}{\sin t \cos t} \][/tex]
We see that LHS = RHS:
[tex]\[ = \tan t + 2 \cot t \][/tex]
Thus, the identity is verified.
### Identity 3: [tex]\((\tan u + \cot u)(\cos u + \sin u) = \csc u + \sec u\)[/tex]
1. Expand the left-hand side (LHS):
[tex]\[ (\tan u + \cot u)(\cos u + \sin u) \][/tex]
2. Distribute:
[tex]\[ = \tan u \cos u + \tan u \sin u + \cot u \cos u + \cot u \sin u \][/tex]
3. Use definitions: [tex]\(\tan u = \frac{\sin u}{\cos u}\)[/tex] and [tex]\(\cot u = \frac{\cos u}{\sin u}\)[/tex]:
[tex]\[ = \frac{\sin u}{\cos u} \cos u + \frac{\sin u}{\cos u} \sin u + \frac{\cos u}{\sin u} \cos u + \frac{\cos u}{\sin u} \sin u \][/tex]
4. Simplify fractions:
[tex]\[ = \sin u + \frac{\sin^2 u}{\cos u} + \cos u + \frac{\cos^2 u}{\sin u} \][/tex]
5. Combine terms with common denominators:
[tex]\[ = \sin u + \cos u + \frac{\sin^2 u + \cos^2 u}{\sin u \cos u} \][/tex]
6. Use the Pythagorean identity [tex]\(\sin^2 u + \cos^2 u = 1\)[/tex]:
[tex]\[ = \sin u + \cos u + \frac{1}{\sin u \cos u} \][/tex]
7. Split the final term:
[tex]\[ = \sin u + \cos u + \csc u + \sec u \][/tex]
Thus, LHS equals the RHS, verifying the identity.
### Identity 4: [tex]\(\tan^2 \alpha - \sin^2 \alpha = \tan^2 \alpha \sin^2 \alpha\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \tan^2 \alpha - \sin^2 \alpha \][/tex]
2. Recall that [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex], so [tex]\(\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}\)[/tex]:
[tex]\[ = \frac{\sin^2 \alpha}{\cos^2 \alpha} - \sin^2 \alpha \][/tex]
3. Use common denominator:
[tex]\[ = \frac{\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha} \][/tex]
4. Factor out [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ = \sin^2 \alpha \left(\frac{1 - \cos^2 \alpha}{\cos^2 \alpha}\right) \][/tex]
5. Use the Pythagorean identity [tex]\(1 - \cos^2 \alpha = \sin^2 \alpha\)[/tex]:
[tex]\[ = \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\cos^2 \alpha}\right) \][/tex]
6. Which simplifies to:
[tex]\[ = \tan^2 \alpha \sin^2 \alpha \][/tex]
Thus, LHS equals RHS, verifying the identity.
### Identity 5: [tex]\(\frac{1 + \sec 4x}{\sin 4x + \tan 4x} = \csc 4x\)[/tex]
1. Start with the left-hand side (LHS):
[tex]\[ \frac{1 + \sec 4x}{\sin 4x + \tan 4x} \][/tex]
2. Recall [tex]\(\sec 4x = \frac{1}{\cos 4x}\)[/tex] and [tex]\(\tan 4x = \frac{\sin 4x}{\cos 4x}\)[/tex]:
[tex]\[ = \frac{1 + \frac{1}{\cos 4x}}{\sin 4x + \frac{\sin 4x}{\cos 4x}} \][/tex]
3. Simplify numerator and denominator:
[tex]\[ = \frac{\frac{\cos 4x + 1}{\cos 4x}}{\frac{\sin 4x \cos 4x + \sin 4x}{\cos 4x}} \][/tex]
4. Combine:
[tex]\[ = \frac{\cos 4x + 1}{\sin 4x (\cos 4x + 1)} \][/tex]
5. Cancel the common term [tex]\(\cos 4x + 1\)[/tex]:
[tex]\[ = \frac{1}{\sin 4x} \][/tex]
6. Recall that [tex]\(\frac{1}{\sin 4x} = \csc 4x\)[/tex]:
[tex]\[ = \csc 4x \][/tex]
Thus, LHS equals RHS, verifying the identity.
All the given identities have been verified correctly.
