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Sagot :
Sure, let's prove the given trigonometric identities step by step.
The given question asks us to prove that:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} = \frac{\cos{A}}{1 + \sin{A}} \][/tex]
### Step 1: Prove \(\tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}}\)
Let's start with the left-hand side of the equation:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) \][/tex]
Using the tangent subtraction formula:
[tex]\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \][/tex]
Let \(x = \frac{\pi}{4}\) and \(y = \frac{A}{2}\). We know that:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1 \][/tex]
So,
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \tan\left(\frac{A}{2}\right)}{1 + \tan\left(\frac{A}{2}\right)} \][/tex]
Next, let's use the half-angle identities for tangent:
[tex]\[ \tan\left(\frac{A}{2}\right) = \frac{\sin A}{1 + \cos A} \quad \text{or} \quad \frac{1 - \cos A}{\sin A} \][/tex]
Using another transformation:
[tex]\[ \tan\left(\frac{A}{2}\right) = \frac{\sin A}{\cos A + \sin A} \][/tex]
Thus, our equation becomes:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \frac{\sin A}{\cos A + \sin A}}{1 + \frac{\sin A}{\cos A + \sin A}} \][/tex]
Simplify the fractions:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{(\cos A + \sin A - \sin A) / (\cos A + \sin A)}{(\cos A + \sin A + \sin A) / (\cos A + \sin A)} \][/tex]
[tex]\[ = \frac{\cos A}{\cos A + \sin A} \][/tex]
Next, we need to reconcile this with:
[tex]\[ \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \][/tex]
Using \(u = \frac{A}{2}\):
[tex]\[ \frac{\cos u - \sin u}{\cos u + \sin u} \][/tex]
We see that it matches. Thus, we have:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \quad \text{proved true} \][/tex]
### Step 2: Prove \(\frac{\cos \frac{A}{2} - \sin \frac{A}{2}}{\cos \frac{A}{2} + \sin \frac{A}{2}} = \frac{\cos A}{1 + \sin A}\)
Starting from the right-hand side of the first equality:
[tex]\[ \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \][/tex]
We want to show this is equal to:
[tex]\[ \frac{\cos A}{1 + \sin A} \][/tex]
We will use the double-angle identities:
[tex]\[ \cos A = \cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right) \][/tex]
\(\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)\) can be rewritten and simplified:
Let's simplify the given:
Divide numerator and denominator by \(\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)\):
[tex]\[ \cos A = \frac{\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)}{1 + 2 \cos \left(\frac{A}{2}\right) \sin \left(\frac{A}{2}\right)} = \frac{\cos \left(\frac{A}{2} \right)^2 - \sin^2 \left(\frac{A}{2}\right)}{1 + 2 \sin\left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)^2} =\cos A /(1+\sin A) \][/tex]
Therefore, this final identity doesn't hold true in every case. Thus:
[tex]\[ \quad \text{proved false} \][/tex]
The given question asks us to prove that:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} = \frac{\cos{A}}{1 + \sin{A}} \][/tex]
### Step 1: Prove \(\tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}}\)
Let's start with the left-hand side of the equation:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) \][/tex]
Using the tangent subtraction formula:
[tex]\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \][/tex]
Let \(x = \frac{\pi}{4}\) and \(y = \frac{A}{2}\). We know that:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1 \][/tex]
So,
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \tan\left(\frac{A}{2}\right)}{1 + \tan\left(\frac{A}{2}\right)} \][/tex]
Next, let's use the half-angle identities for tangent:
[tex]\[ \tan\left(\frac{A}{2}\right) = \frac{\sin A}{1 + \cos A} \quad \text{or} \quad \frac{1 - \cos A}{\sin A} \][/tex]
Using another transformation:
[tex]\[ \tan\left(\frac{A}{2}\right) = \frac{\sin A}{\cos A + \sin A} \][/tex]
Thus, our equation becomes:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \frac{\sin A}{\cos A + \sin A}}{1 + \frac{\sin A}{\cos A + \sin A}} \][/tex]
Simplify the fractions:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{(\cos A + \sin A - \sin A) / (\cos A + \sin A)}{(\cos A + \sin A + \sin A) / (\cos A + \sin A)} \][/tex]
[tex]\[ = \frac{\cos A}{\cos A + \sin A} \][/tex]
Next, we need to reconcile this with:
[tex]\[ \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \][/tex]
Using \(u = \frac{A}{2}\):
[tex]\[ \frac{\cos u - \sin u}{\cos u + \sin u} \][/tex]
We see that it matches. Thus, we have:
[tex]\[ \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \quad \text{proved true} \][/tex]
### Step 2: Prove \(\frac{\cos \frac{A}{2} - \sin \frac{A}{2}}{\cos \frac{A}{2} + \sin \frac{A}{2}} = \frac{\cos A}{1 + \sin A}\)
Starting from the right-hand side of the first equality:
[tex]\[ \frac{\cos\frac{A}{2} - \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}} \][/tex]
We want to show this is equal to:
[tex]\[ \frac{\cos A}{1 + \sin A} \][/tex]
We will use the double-angle identities:
[tex]\[ \cos A = \cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right) \][/tex]
\(\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)\) can be rewritten and simplified:
Let's simplify the given:
Divide numerator and denominator by \(\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)\):
[tex]\[ \cos A = \frac{\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)}{1 + 2 \cos \left(\frac{A}{2}\right) \sin \left(\frac{A}{2}\right)} = \frac{\cos \left(\frac{A}{2} \right)^2 - \sin^2 \left(\frac{A}{2}\right)}{1 + 2 \sin\left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)^2} =\cos A /(1+\sin A) \][/tex]
Therefore, this final identity doesn't hold true in every case. Thus:
[tex]\[ \quad \text{proved false} \][/tex]
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