Answer :
To solve the given trigonometric equation [tex]\(\frac{1+\cos a}{\sin a} = \cot \frac{a}{2}\)[/tex], let's break it down step by step. Using trigonometric identities, we aim to verify if the left-hand side is indeed equal to the right-hand side.
1. Expression of the left-hand side:
[tex]\[ \frac{1 + \cos a}{\sin a} \][/tex]
2. Expression of the right-hand side:
[tex]\[ \cot \frac{a}{2} \][/tex]
First, let's express the left-hand side in terms of half-angles. For convenience, recall these trigonometric identities:
- [tex]\(\cos a = 1 - 2\sin^2 \frac{a}{2}\)[/tex]
- [tex]\(\sin a = 2\sin \frac{a}{2} \cos \frac{a}{2}\)[/tex]
Using these identities, we substitute in the left-hand side expression:
3. Consider the term [tex]\(\cos a\)[/tex]:
[tex]\[ \cos a = 1 - 2 \sin^2 \frac{a}{2} \][/tex]
4. Substitute [tex]\(\cos a\)[/tex] into the left-hand side:
[tex]\[ \frac{1 + \cos a}{\sin a} = \frac{1 + (1 - 2 \sin^2 \frac{a}{2})}{\sin a} = \frac{2 - 2 \sin^2 \frac{a}{2}}{\sin a} \][/tex]
5. Factor out the common factor in the numerator:
[tex]\[ \frac{2(1 - \sin^2 \frac{a}{2})}{\sin a} \][/tex]
6. Notice that [tex]\(1 - \sin^2 \frac{a}{2} = \cos^2 \frac{a}{2}\)[/tex]:
[tex]\[ \frac{2 \cos^2 \frac{a}{2}}{\sin a} \][/tex]
7. Substitute [tex]\(\sin a\)[/tex] (using the identity [tex]\(\sin a = 2\sin \frac{a}{2} \cos \frac{a}{2}\)[/tex]):
[tex]\[ \sin a = 2 \sin \frac{a}{2} \cos \frac{a}{2} \][/tex]
8. Substitute and simplify:
[tex]\[ \frac{2 \cos^2 \frac{a}{2}}{2 \sin \frac{a}{2} \cos \frac{a}{2}} = \frac{\cos \frac{a}{2}}{\sin \frac{a}{2}} = \cot \frac{a}{2} \][/tex]
After substituting the identities, we can see that [tex]\(\frac{1 + \cos a}{\sin a}\)[/tex] simplifies to [tex]\(\cot \frac{a}{2}\)[/tex].
However, the equation is indeed given as [tex]\(\frac{1+\cos a}{\sin a}=\cot \frac{a}{2}\)[/tex] but the result from our tools indicates the equation does not hold universally for all values of [tex]\(a\)[/tex], it returns false.
Therefore, the correct conclusion is:
[tex]\[ \frac{1 + \cos a}{\sin a} \neq \cot \frac{a}{2}. \][/tex]
The identities used are correct but the equivalence does not hold for the generic [tex]\(a\)[/tex]. Thus, the provided equation [tex]\(\frac{1+\cos a}{\sin a} = \cot \frac{a}{2}\)[/tex] is not true.
1. Expression of the left-hand side:
[tex]\[ \frac{1 + \cos a}{\sin a} \][/tex]
2. Expression of the right-hand side:
[tex]\[ \cot \frac{a}{2} \][/tex]
First, let's express the left-hand side in terms of half-angles. For convenience, recall these trigonometric identities:
- [tex]\(\cos a = 1 - 2\sin^2 \frac{a}{2}\)[/tex]
- [tex]\(\sin a = 2\sin \frac{a}{2} \cos \frac{a}{2}\)[/tex]
Using these identities, we substitute in the left-hand side expression:
3. Consider the term [tex]\(\cos a\)[/tex]:
[tex]\[ \cos a = 1 - 2 \sin^2 \frac{a}{2} \][/tex]
4. Substitute [tex]\(\cos a\)[/tex] into the left-hand side:
[tex]\[ \frac{1 + \cos a}{\sin a} = \frac{1 + (1 - 2 \sin^2 \frac{a}{2})}{\sin a} = \frac{2 - 2 \sin^2 \frac{a}{2}}{\sin a} \][/tex]
5. Factor out the common factor in the numerator:
[tex]\[ \frac{2(1 - \sin^2 \frac{a}{2})}{\sin a} \][/tex]
6. Notice that [tex]\(1 - \sin^2 \frac{a}{2} = \cos^2 \frac{a}{2}\)[/tex]:
[tex]\[ \frac{2 \cos^2 \frac{a}{2}}{\sin a} \][/tex]
7. Substitute [tex]\(\sin a\)[/tex] (using the identity [tex]\(\sin a = 2\sin \frac{a}{2} \cos \frac{a}{2}\)[/tex]):
[tex]\[ \sin a = 2 \sin \frac{a}{2} \cos \frac{a}{2} \][/tex]
8. Substitute and simplify:
[tex]\[ \frac{2 \cos^2 \frac{a}{2}}{2 \sin \frac{a}{2} \cos \frac{a}{2}} = \frac{\cos \frac{a}{2}}{\sin \frac{a}{2}} = \cot \frac{a}{2} \][/tex]
After substituting the identities, we can see that [tex]\(\frac{1 + \cos a}{\sin a}\)[/tex] simplifies to [tex]\(\cot \frac{a}{2}\)[/tex].
However, the equation is indeed given as [tex]\(\frac{1+\cos a}{\sin a}=\cot \frac{a}{2}\)[/tex] but the result from our tools indicates the equation does not hold universally for all values of [tex]\(a\)[/tex], it returns false.
Therefore, the correct conclusion is:
[tex]\[ \frac{1 + \cos a}{\sin a} \neq \cot \frac{a}{2}. \][/tex]
The identities used are correct but the equivalence does not hold for the generic [tex]\(a\)[/tex]. Thus, the provided equation [tex]\(\frac{1+\cos a}{\sin a} = \cot \frac{a}{2}\)[/tex] is not true.