Answer :
To determine which of the given expressions are equivalent to [tex]\(\cos \varnothing\)[/tex], we will simplify each expression step-by-step.
### Expression (a)
[tex]\[\sin \varnothing \cos \varnothing (\tan \varnothing + \cot \varnothing)\][/tex]
1. Simplify the terms inside the parentheses:
- [tex]\(\tan \varnothing = \frac{\sin \varnothing}{\cos \varnothing}\)[/tex]
- [tex]\(\cot \varnothing = \frac{\cos \varnothing}{\sin \varnothing}\)[/tex]
2. Adding these together:
[tex]\[ \tan \varnothing + \cot \varnothing = \frac{\sin \varnothing}{\cos \varnothing} + \frac{\cos \varnothing}{\sin \varnothing} = \frac{\sin^2 \varnothing + \cos^2 \varnothing}{\sin \varnothing \cos \varnothing} \][/tex]
3. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we get:
[tex]\[ \tan \varnothing + \cot \varnothing = \frac{1}{\sin \varnothing \cos \varnothing} \][/tex]
4. Thus the entire expression becomes:
[tex]\[ \sin \varnothing \cos \varnothing \times \frac{1}{\sin \varnothing \cos \varnothing} = 1 \][/tex]
Since [tex]\(1 \neq \cos \varnothing\)[/tex], expression (a) is not equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (b)
[tex]\[\sec \varnothing - \tan \emptyset \sin \varnothing\][/tex]
1. Rewrite in terms of sine and cosine:
- [tex]\(\sec \varnothing = \frac{1}{\cos \varnothing}\)[/tex]
- [tex]\(\tan \varnothing = \frac{\sin \varnothing}{\cos \varnothing}\)[/tex]
2. Substituting these, we get:
[tex]\[ \sec \varnothing - \tan \varnothing \sin \varnothing = \frac{1}{\cos \varnothing} - \frac{\sin \varnothing}{\cos \varnothing} \sin \varnothing \][/tex]
3. Simplify the expression:
[tex]\[ \frac{1}{\cos \varnothing} - \frac{\sin^2 \varnothing}{\cos \varnothing} = \frac{1 - \sin^2 \varnothing}{\cos \varnothing} \][/tex]
4. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we have:
[tex]\[ 1 - \sin^2 \varnothing = \cos^2 \varnothing \][/tex]
5. Thus, the expression becomes:
[tex]\[ \frac{\cos^2 \varnothing}{\cos \varnothing} = \cos \varnothing \][/tex]
Expression (b) is equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (c)
[tex]\[\frac{\cos^2 \varnothing}{1 - \sin \varnothing}\][/tex]
This expression does not simplify directly to [tex]\(\cos \varnothing\)[/tex]. So, expression (c) is not equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (d)
[tex]\[\sec \varnothing (1 - \sin^2 \varnothing)\][/tex]
1. Rewrite in terms of sine and cosine:
- [tex]\(\sec \varnothing = \frac{1}{\cos \varnothing}\)[/tex]
2. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we have:
[tex]\[ 1 - \sin^2 \varnothing = \cos^2 \varnothing \][/tex]
3. Substituting these, we get:
[tex]\[ \frac{1}{\cos \varnothing} \times \cos^2 \varnothing = \cos \varnothing \][/tex]
Expression (d) is equivalent to [tex]\(\cos \varnothing\)[/tex].
### Conclusion
The expressions that are equivalent to [tex]\(\cos \varnothing\)[/tex] are:
- Expression (b)
- Expression (d)
The correct answers are expressions (b) and (d).
### Expression (a)
[tex]\[\sin \varnothing \cos \varnothing (\tan \varnothing + \cot \varnothing)\][/tex]
1. Simplify the terms inside the parentheses:
- [tex]\(\tan \varnothing = \frac{\sin \varnothing}{\cos \varnothing}\)[/tex]
- [tex]\(\cot \varnothing = \frac{\cos \varnothing}{\sin \varnothing}\)[/tex]
2. Adding these together:
[tex]\[ \tan \varnothing + \cot \varnothing = \frac{\sin \varnothing}{\cos \varnothing} + \frac{\cos \varnothing}{\sin \varnothing} = \frac{\sin^2 \varnothing + \cos^2 \varnothing}{\sin \varnothing \cos \varnothing} \][/tex]
3. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we get:
[tex]\[ \tan \varnothing + \cot \varnothing = \frac{1}{\sin \varnothing \cos \varnothing} \][/tex]
4. Thus the entire expression becomes:
[tex]\[ \sin \varnothing \cos \varnothing \times \frac{1}{\sin \varnothing \cos \varnothing} = 1 \][/tex]
Since [tex]\(1 \neq \cos \varnothing\)[/tex], expression (a) is not equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (b)
[tex]\[\sec \varnothing - \tan \emptyset \sin \varnothing\][/tex]
1. Rewrite in terms of sine and cosine:
- [tex]\(\sec \varnothing = \frac{1}{\cos \varnothing}\)[/tex]
- [tex]\(\tan \varnothing = \frac{\sin \varnothing}{\cos \varnothing}\)[/tex]
2. Substituting these, we get:
[tex]\[ \sec \varnothing - \tan \varnothing \sin \varnothing = \frac{1}{\cos \varnothing} - \frac{\sin \varnothing}{\cos \varnothing} \sin \varnothing \][/tex]
3. Simplify the expression:
[tex]\[ \frac{1}{\cos \varnothing} - \frac{\sin^2 \varnothing}{\cos \varnothing} = \frac{1 - \sin^2 \varnothing}{\cos \varnothing} \][/tex]
4. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we have:
[tex]\[ 1 - \sin^2 \varnothing = \cos^2 \varnothing \][/tex]
5. Thus, the expression becomes:
[tex]\[ \frac{\cos^2 \varnothing}{\cos \varnothing} = \cos \varnothing \][/tex]
Expression (b) is equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (c)
[tex]\[\frac{\cos^2 \varnothing}{1 - \sin \varnothing}\][/tex]
This expression does not simplify directly to [tex]\(\cos \varnothing\)[/tex]. So, expression (c) is not equivalent to [tex]\(\cos \varnothing\)[/tex].
### Expression (d)
[tex]\[\sec \varnothing (1 - \sin^2 \varnothing)\][/tex]
1. Rewrite in terms of sine and cosine:
- [tex]\(\sec \varnothing = \frac{1}{\cos \varnothing}\)[/tex]
2. Using the Pythagorean identity [tex]\(\sin^2 \varnothing + \cos^2 \varnothing = 1\)[/tex], we have:
[tex]\[ 1 - \sin^2 \varnothing = \cos^2 \varnothing \][/tex]
3. Substituting these, we get:
[tex]\[ \frac{1}{\cos \varnothing} \times \cos^2 \varnothing = \cos \varnothing \][/tex]
Expression (d) is equivalent to [tex]\(\cos \varnothing\)[/tex].
### Conclusion
The expressions that are equivalent to [tex]\(\cos \varnothing\)[/tex] are:
- Expression (b)
- Expression (d)
The correct answers are expressions (b) and (d).
