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Sagot :
Answer:
The proof is done in the step-by-step explanation below.
Step-by-step explanation:
We are given the following identity:
[tex]\frac{\sin{A}\tan{A}}{1-\cos{A}}[/tex]
And we have to show that this is equals to:
[tex]1 + \sec{A}[/tex]
Multiplying numerator and denominator by the conjugate of the denominator:
[tex]\frac{\sin{A}\tan{A}}{1-\cos{A}} \times \frac{1+\cos{A}}{1+\cos{A}}[/tex]
[tex]\frac{\sin{A}\tan{A}(1+\cos{A})}{1 - \cos^2{A}}[/tex]
We use these following identities:
[tex]\sin^2{A} + \cos^2{A} = 1[/tex]
So
[tex]1 - \cos^2{A} = \sin^2{A}[/tex]
Also:
[tex]\tan{A} = \frac{\sin{A}}{\cos{A}}[/tex]
Then
[tex]\frac{\sin{A}\sin{A}(1+\cos{A})}{\cos{A}\sin^2{A}}[/tex]
[tex]\frac{\sin^2{A}(1+\cos{A})}{\cos{A}\sin^2{A}}[/tex]
[tex]\frac{1 + \cos{A}}{\cos{A}}[/tex]
[tex]\frac{1}{\cos{A}} + \frac{\cos{A}}{\cos{A}}[/tex]
[tex]\frac{1}{\cos{A}} + 1[/tex]
Since:
[tex]\sec{A} = \frac{1}{\cos{A}}[/tex]
We have that:
[tex]1 + \sec{A}[/tex]
Thus, the proof is done.
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