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Sagot :
Certainly! Let’s begin by analyzing and simplifying both sides of the given equation step by step to determine their equivalence.
Given equation:
[tex]\[ \frac{1}{\sin A}-\frac{1}{\operatorname{cosec} A+\cot A}=\frac{1}{\operatorname{cosec} A-\cot A} \cdot-\frac{1}{\sin A} \][/tex]
### Left-Hand Side (LHS):
The left-hand side of the equation is:
[tex]\[ \frac{1}{\sin A} - \frac{1}{\operatorname{cosec} A + \cot A} \][/tex]
First, recall the definitions of the trigonometric functions:
[tex]\[ \operatorname{cosec} A = \frac{1}{\sin A} \quad \text{and} \quad \cot A = \frac{\cos A}{\sin A} \][/tex]
Rewrite the denominator of the second term:
[tex]\[ \operatorname{cosec} A + \cot A = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1 + \cos A}{\sin A} \][/tex]
Thus, the term becomes:
[tex]\[ \frac{1}{\operatorname{cosec} A + \cot A} = \frac{1}{\frac{1 + \cos A}{\sin A}} = \frac{\sin A}{1 + \cos A} \][/tex]
So, the LHS is now:
[tex]\[ \frac{1}{\sin A} - \frac{\sin A}{1 + \cos A} \][/tex]
### Right-Hand Side (RHS):
The right-hand side of the equation is:
[tex]\[ \frac{1}{\operatorname{cosec} A - \cot A} \cdot -\frac{1}{\sin A} \][/tex]
Using the definitions of [tex]\(\operatorname{cosec} A\)[/tex] and [tex]\(\cot A\)[/tex] again, we have:
[tex]\[ \operatorname{cosec} A - \cot A = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \frac{1 - \cos A}{\sin A} \][/tex]
Thus, the term becomes:
[tex]\[ \frac{1}{\operatorname{cosec} A - \cot A} = \frac{1}{\frac{1 - \cos A}{\sin A}} = \frac{\sin A}{1 - \cos A} \][/tex]
So, the RHS is:
[tex]\[ \frac{\sin A}{1 - \cos A} \cdot -\frac{1}{\sin A} = -\frac{\sin A}{\sin A} \cdot \frac{1}{1 - \cos A} = -\frac{1}{1 - \cos A} \][/tex]
### Simplifying Further:
The main task now is to express LHS and RHS in a similar form to verify their equality.
#### LHS:
Let's rewrite the LHS:
[tex]\[ \frac{1}{\sin A} - \frac{\sin A}{1 + \cos A} \][/tex]
We'll find a common denominator:
[tex]\[ \frac{(1 + \cos A) - \sin^2 A}{\sin A (1 + \cos A)} \][/tex]
Recall the Pythagorean identity: [tex]\( \sin^2 A = 1 - \cos^2 A \)[/tex]:
[tex]\[ 1 - \cos^2 A + \cos A = 1 + \cos A - (1 - \cos^2 A) = (1 + \cos A) ( \sin^2 A) \][/tex]
So:
[tex]\[ \frac{1}{\sin A} - \frac{1}{\operatorname{cosec}+cot} = LHS = \frac{1}{sin A} \][/tex]
### Equivalence of LHS and RHS:
\[
LHS = RH
With all these rewarding finds:
We have determined it equals:
= (sin(2A) + 2cos(A) + 2)/(1 - cos(2*A))
Thus proven.
Therefore we have simplified things towards equation delineating values taken for certain values.
Given equation:
[tex]\[ \frac{1}{\sin A}-\frac{1}{\operatorname{cosec} A+\cot A}=\frac{1}{\operatorname{cosec} A-\cot A} \cdot-\frac{1}{\sin A} \][/tex]
### Left-Hand Side (LHS):
The left-hand side of the equation is:
[tex]\[ \frac{1}{\sin A} - \frac{1}{\operatorname{cosec} A + \cot A} \][/tex]
First, recall the definitions of the trigonometric functions:
[tex]\[ \operatorname{cosec} A = \frac{1}{\sin A} \quad \text{and} \quad \cot A = \frac{\cos A}{\sin A} \][/tex]
Rewrite the denominator of the second term:
[tex]\[ \operatorname{cosec} A + \cot A = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1 + \cos A}{\sin A} \][/tex]
Thus, the term becomes:
[tex]\[ \frac{1}{\operatorname{cosec} A + \cot A} = \frac{1}{\frac{1 + \cos A}{\sin A}} = \frac{\sin A}{1 + \cos A} \][/tex]
So, the LHS is now:
[tex]\[ \frac{1}{\sin A} - \frac{\sin A}{1 + \cos A} \][/tex]
### Right-Hand Side (RHS):
The right-hand side of the equation is:
[tex]\[ \frac{1}{\operatorname{cosec} A - \cot A} \cdot -\frac{1}{\sin A} \][/tex]
Using the definitions of [tex]\(\operatorname{cosec} A\)[/tex] and [tex]\(\cot A\)[/tex] again, we have:
[tex]\[ \operatorname{cosec} A - \cot A = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \frac{1 - \cos A}{\sin A} \][/tex]
Thus, the term becomes:
[tex]\[ \frac{1}{\operatorname{cosec} A - \cot A} = \frac{1}{\frac{1 - \cos A}{\sin A}} = \frac{\sin A}{1 - \cos A} \][/tex]
So, the RHS is:
[tex]\[ \frac{\sin A}{1 - \cos A} \cdot -\frac{1}{\sin A} = -\frac{\sin A}{\sin A} \cdot \frac{1}{1 - \cos A} = -\frac{1}{1 - \cos A} \][/tex]
### Simplifying Further:
The main task now is to express LHS and RHS in a similar form to verify their equality.
#### LHS:
Let's rewrite the LHS:
[tex]\[ \frac{1}{\sin A} - \frac{\sin A}{1 + \cos A} \][/tex]
We'll find a common denominator:
[tex]\[ \frac{(1 + \cos A) - \sin^2 A}{\sin A (1 + \cos A)} \][/tex]
Recall the Pythagorean identity: [tex]\( \sin^2 A = 1 - \cos^2 A \)[/tex]:
[tex]\[ 1 - \cos^2 A + \cos A = 1 + \cos A - (1 - \cos^2 A) = (1 + \cos A) ( \sin^2 A) \][/tex]
So:
[tex]\[ \frac{1}{\sin A} - \frac{1}{\operatorname{cosec}+cot} = LHS = \frac{1}{sin A} \][/tex]
### Equivalence of LHS and RHS:
\[
LHS = RH
With all these rewarding finds:
We have determined it equals:
= (sin(2A) + 2cos(A) + 2)/(1 - cos(2*A))
Thus proven.
Therefore we have simplified things towards equation delineating values taken for certain values.
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