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Sagot :
Sure, let us verify the given equation step-by-step:
[tex]\[ \frac{(1+\cos \beta)+(1-\cos \beta)^2}{\sin^2 \beta} = 2\left(1+2 \cot^2 \beta\right) \][/tex]
### Step-by-Step Solution:
1. Simplify the Left-Hand Side (LHS):
[tex]\[ \frac{(1+\cos \beta)+(1-\cos \beta)^2}{\sin^2 \beta} \][/tex]
Start with simplifying the numerator:
[tex]\[ (1+\cos \beta) + (1 - \cos \beta)^2 \][/tex]
Expand [tex]\((1 - \cos \beta)^2\)[/tex]:
[tex]\[ (1 - \cos \beta)^2 = 1 - 2\cos \beta + \cos^2 \beta \][/tex]
Therefore, the numerator becomes:
[tex]\[ (1 + \cos \beta) + 1 - 2\cos \beta + \cos^2 \beta \][/tex]
Combine like terms:
[tex]\[ 1 + \cos \beta + 1 - 2\cos \beta + \cos^2 \beta = \cos^2 \beta - \cos \beta + 2 \][/tex]
Now, let's write the entire fraction:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \][/tex]
2. Simplify the Right-Hand Side (RHS):
[tex]\[ 2 \left(1 + 2 \cot^2 \beta\right) \][/tex]
Recall that:
[tex]\[ \cot \beta = \frac{\cos \beta}{\sin \beta} \][/tex]
Therefore:
[tex]\[ \cot^2 \beta = \left(\frac{\cos \beta}{\sin \beta}\right)^2 = \frac{\cos^2 \beta}{\sin^2 \beta} \][/tex]
Substitute [tex]\(\cot^2 \beta\)[/tex] back into the equation:
[tex]\[ 2 \left(1 + 2 \frac{\cos^2 \beta}{\sin^2 \beta}\right) \][/tex]
Distribute the 2:
[tex]\[ 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} = 2 + 4 \cot^2 \beta \][/tex]
3. Compare LHS and RHS:
LHS:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \][/tex]
RHS:
[tex]\[ 2 + 4 \cot^2 \beta = 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} \][/tex]
Rewrite RHS with the common denominator [tex]\(\sin^2 \beta \)[/tex]:
[tex]\[ 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} = \frac{2 \sin^2 \beta + 4 \cos^2 \beta}{\sin^2 \beta} \][/tex]
Now, the comparison is between:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \quad \text{and} \quad \frac{2 \sin^2 \beta + 4 \cos^2 \beta}{\sin^2 \beta} \][/tex]
4. Conclusion:
Upon simplification, the fractions on both sides are not equal, and therefore:
[tex]\[ \frac{(\cos^2 \beta - \cos \beta + 2)}{\sin^2 \beta} \neq 2 + 4 \cot^2 \beta \][/tex]
Hence, the given equation is not an identity. The left-hand side does not equate to the right-hand side for all values of [tex]\(\beta\)[/tex].
[tex]\[ \frac{(1+\cos \beta)+(1-\cos \beta)^2}{\sin^2 \beta} = 2\left(1+2 \cot^2 \beta\right) \][/tex]
### Step-by-Step Solution:
1. Simplify the Left-Hand Side (LHS):
[tex]\[ \frac{(1+\cos \beta)+(1-\cos \beta)^2}{\sin^2 \beta} \][/tex]
Start with simplifying the numerator:
[tex]\[ (1+\cos \beta) + (1 - \cos \beta)^2 \][/tex]
Expand [tex]\((1 - \cos \beta)^2\)[/tex]:
[tex]\[ (1 - \cos \beta)^2 = 1 - 2\cos \beta + \cos^2 \beta \][/tex]
Therefore, the numerator becomes:
[tex]\[ (1 + \cos \beta) + 1 - 2\cos \beta + \cos^2 \beta \][/tex]
Combine like terms:
[tex]\[ 1 + \cos \beta + 1 - 2\cos \beta + \cos^2 \beta = \cos^2 \beta - \cos \beta + 2 \][/tex]
Now, let's write the entire fraction:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \][/tex]
2. Simplify the Right-Hand Side (RHS):
[tex]\[ 2 \left(1 + 2 \cot^2 \beta\right) \][/tex]
Recall that:
[tex]\[ \cot \beta = \frac{\cos \beta}{\sin \beta} \][/tex]
Therefore:
[tex]\[ \cot^2 \beta = \left(\frac{\cos \beta}{\sin \beta}\right)^2 = \frac{\cos^2 \beta}{\sin^2 \beta} \][/tex]
Substitute [tex]\(\cot^2 \beta\)[/tex] back into the equation:
[tex]\[ 2 \left(1 + 2 \frac{\cos^2 \beta}{\sin^2 \beta}\right) \][/tex]
Distribute the 2:
[tex]\[ 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} = 2 + 4 \cot^2 \beta \][/tex]
3. Compare LHS and RHS:
LHS:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \][/tex]
RHS:
[tex]\[ 2 + 4 \cot^2 \beta = 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} \][/tex]
Rewrite RHS with the common denominator [tex]\(\sin^2 \beta \)[/tex]:
[tex]\[ 2 + 4 \frac{\cos^2 \beta}{\sin^2 \beta} = \frac{2 \sin^2 \beta + 4 \cos^2 \beta}{\sin^2 \beta} \][/tex]
Now, the comparison is between:
[tex]\[ \frac{\cos^2 \beta - \cos \beta + 2}{\sin^2 \beta} \quad \text{and} \quad \frac{2 \sin^2 \beta + 4 \cos^2 \beta}{\sin^2 \beta} \][/tex]
4. Conclusion:
Upon simplification, the fractions on both sides are not equal, and therefore:
[tex]\[ \frac{(\cos^2 \beta - \cos \beta + 2)}{\sin^2 \beta} \neq 2 + 4 \cot^2 \beta \][/tex]
Hence, the given equation is not an identity. The left-hand side does not equate to the right-hand side for all values of [tex]\(\beta\)[/tex].
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