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Sagot :
To solve the given trigonometric expression [tex]\((\operatorname{cosec} A - \sin A) \cdot (\sec A - \cos A) \cdot (\tan A + \cot A)\)[/tex], let's proceed step-by-step to understand how we can simplify and solve it.
First, let's recall the definitions and basic identities of the trigonometric functions involved:
1. [tex]\(\operatorname{cosec} A = \frac{1}{\sin A}\)[/tex]
2. [tex]\(\sec A = \frac{1}{\cos A}\)[/tex]
3. [tex]\(\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}\)[/tex]
Let's break down the expression into its three main parts:
### Part 1: [tex]\(\operatorname{cosec} A - \sin A\)[/tex]
[tex]\[ \operatorname{cosec} A - \sin A = \frac{1}{\sin A} - \sin A \][/tex]
### Part 2: [tex]\(\sec A - \cos A\)[/tex]
[tex]\[ \sec A - \cos A = \frac{1}{\cos A} - \cos A \][/tex]
### Part 3: [tex]\(\tan A + \cot A\)[/tex]
[tex]\[ \tan A + \cot A = \tan A + \frac{1}{\tan A} \][/tex]
Now we combine all these simplified forms into the main expression:
[tex]\[ \left( \frac{1}{\sin A} - \sin A \right) \cdot \left( \frac{1}{\cos A} - \cos A \right) \cdot \left( \tan A + \frac{\cos A}{\sin A} \right) \][/tex]
We can further simplify [tex]\(\tan A + \frac{\cos A}{\sin A}\)[/tex]:
[tex]\[ \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} \][/tex]
Using the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Therefore:
[tex]\[ \tan A + \cot A = \frac{1}{\sin A \cos A} \][/tex]
Now substitute this back into our main expression:
[tex]\[ \left( \frac{1}{\sin A} - \sin A \right) \cdot \left( \frac{1}{\cos A} - \cos A \right) \cdot \frac{1}{\sin A \cos A} \][/tex]
### Combining All Parts:
1. Consider [tex]\(\left( \frac{1}{\sin A} - \sin A \right)\)[/tex]:
[tex]\[ \frac{1}{\sin A} - \sin A = \frac{1 - \sin^2 A}{\sin A} = \frac{\cos^2 A}{\sin A} \][/tex]
Thus:
[tex]\[ \frac{\cos^2 A}{\sin A} \][/tex]
2. Consider [tex]\(\left( \frac{1}{\cos A} - \cos A \right)\)[/tex]:
[tex]\[ \frac{1}{\cos A} - \cos A = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} \][/tex]
Thus:
[tex]\[ \frac{\sin^2 A}{\cos A} \][/tex]
3. Combining them together we get:
[tex]\[ (\frac{\cos^2 A}{\sin A}) \cdot (\frac{\sin^2 A}{\cos A}) \cdot \frac{1}{\sin A \cos A} \][/tex]
### Simplifying Further:
Combine the fractions:
[tex]\[ \frac{\cos^2 A \cdot \sin^2 A}{\sin A \cos A} \cdot \frac{1}{\sin A \cos A} = \frac{\cos^2 A \sin^2 A}{\sin A \cos A \sin A \cos A} = \frac{\cos^2 A \sin^2 A}{\sin^2 A \cos^2 A} = \frac{1}{1} = 1 \][/tex]
Thus, the given expression simplifies to 1, confirming that:
[tex]\[ (\operatorname{cosec} A - \sin A) \cdot (\sec A - \cos A) \cdot (\tan A + \cot A) = 1 \][/tex]
So, the given trigonometric expression holds true and simplifies to 1 for all [tex]\(A\)[/tex] in the domain of the expression.
First, let's recall the definitions and basic identities of the trigonometric functions involved:
1. [tex]\(\operatorname{cosec} A = \frac{1}{\sin A}\)[/tex]
2. [tex]\(\sec A = \frac{1}{\cos A}\)[/tex]
3. [tex]\(\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}\)[/tex]
Let's break down the expression into its three main parts:
### Part 1: [tex]\(\operatorname{cosec} A - \sin A\)[/tex]
[tex]\[ \operatorname{cosec} A - \sin A = \frac{1}{\sin A} - \sin A \][/tex]
### Part 2: [tex]\(\sec A - \cos A\)[/tex]
[tex]\[ \sec A - \cos A = \frac{1}{\cos A} - \cos A \][/tex]
### Part 3: [tex]\(\tan A + \cot A\)[/tex]
[tex]\[ \tan A + \cot A = \tan A + \frac{1}{\tan A} \][/tex]
Now we combine all these simplified forms into the main expression:
[tex]\[ \left( \frac{1}{\sin A} - \sin A \right) \cdot \left( \frac{1}{\cos A} - \cos A \right) \cdot \left( \tan A + \frac{\cos A}{\sin A} \right) \][/tex]
We can further simplify [tex]\(\tan A + \frac{\cos A}{\sin A}\)[/tex]:
[tex]\[ \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} \][/tex]
Using the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Therefore:
[tex]\[ \tan A + \cot A = \frac{1}{\sin A \cos A} \][/tex]
Now substitute this back into our main expression:
[tex]\[ \left( \frac{1}{\sin A} - \sin A \right) \cdot \left( \frac{1}{\cos A} - \cos A \right) \cdot \frac{1}{\sin A \cos A} \][/tex]
### Combining All Parts:
1. Consider [tex]\(\left( \frac{1}{\sin A} - \sin A \right)\)[/tex]:
[tex]\[ \frac{1}{\sin A} - \sin A = \frac{1 - \sin^2 A}{\sin A} = \frac{\cos^2 A}{\sin A} \][/tex]
Thus:
[tex]\[ \frac{\cos^2 A}{\sin A} \][/tex]
2. Consider [tex]\(\left( \frac{1}{\cos A} - \cos A \right)\)[/tex]:
[tex]\[ \frac{1}{\cos A} - \cos A = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} \][/tex]
Thus:
[tex]\[ \frac{\sin^2 A}{\cos A} \][/tex]
3. Combining them together we get:
[tex]\[ (\frac{\cos^2 A}{\sin A}) \cdot (\frac{\sin^2 A}{\cos A}) \cdot \frac{1}{\sin A \cos A} \][/tex]
### Simplifying Further:
Combine the fractions:
[tex]\[ \frac{\cos^2 A \cdot \sin^2 A}{\sin A \cos A} \cdot \frac{1}{\sin A \cos A} = \frac{\cos^2 A \sin^2 A}{\sin A \cos A \sin A \cos A} = \frac{\cos^2 A \sin^2 A}{\sin^2 A \cos^2 A} = \frac{1}{1} = 1 \][/tex]
Thus, the given expression simplifies to 1, confirming that:
[tex]\[ (\operatorname{cosec} A - \sin A) \cdot (\sec A - \cos A) \cdot (\tan A + \cot A) = 1 \][/tex]
So, the given trigonometric expression holds true and simplifies to 1 for all [tex]\(A\)[/tex] in the domain of the expression.
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