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Sagot :
To simplify the expression
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} - \frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)}, \][/tex]
let's break it down into simpler parts and simplify each one step-by-step.
First, we'll simplify the expression
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} \][/tex]
Recall the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
Also note that the denominator can be simplified using the difference of squares:
[tex]\[ (1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta \][/tex]
So we have:
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{\sin^2 \theta}{1 - \cos^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \][/tex]
Next, let's simplify the expression
[tex]\[ \frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)} \][/tex]
Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \][/tex]
Thus,
[tex]\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Now, simplify the expression in the denominator:
[tex]\[ (\sec \theta - 1)(\sec \theta + 1) \][/tex]
This is again a difference of squares:
[tex]\[ \sec^2 \theta - 1 \][/tex]
Recall the Pythagorean identity for secant and tangent:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \implies \sec^2 \theta - 1 = \tan^2 \theta \][/tex]
Thus, the denominator becomes:
[tex]\[ (\sec^2 \theta - 1) = \tan^2 \theta \][/tex]
Now, substituting back into our expression:
[tex]\[ \frac{\tan^2 \theta}{\tan^2 \theta} = 1 \][/tex]
Finally, we subtract these simplified expressions:
[tex]\[ 1 - 1 = 0 \][/tex]
Therefore, the simplified form of the original expression is:
[tex]\[ 0 \][/tex]
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} - \frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)}, \][/tex]
let's break it down into simpler parts and simplify each one step-by-step.
First, we'll simplify the expression
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} \][/tex]
Recall the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
Also note that the denominator can be simplified using the difference of squares:
[tex]\[ (1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta \][/tex]
So we have:
[tex]\[ \frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{\sin^2 \theta}{1 - \cos^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \][/tex]
Next, let's simplify the expression
[tex]\[ \frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)} \][/tex]
Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \][/tex]
Thus,
[tex]\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Now, simplify the expression in the denominator:
[tex]\[ (\sec \theta - 1)(\sec \theta + 1) \][/tex]
This is again a difference of squares:
[tex]\[ \sec^2 \theta - 1 \][/tex]
Recall the Pythagorean identity for secant and tangent:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \implies \sec^2 \theta - 1 = \tan^2 \theta \][/tex]
Thus, the denominator becomes:
[tex]\[ (\sec^2 \theta - 1) = \tan^2 \theta \][/tex]
Now, substituting back into our expression:
[tex]\[ \frac{\tan^2 \theta}{\tan^2 \theta} = 1 \][/tex]
Finally, we subtract these simplified expressions:
[tex]\[ 1 - 1 = 0 \][/tex]
Therefore, the simplified form of the original expression is:
[tex]\[ 0 \][/tex]
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