Answer:
[tex](\frac{sin\ x}{1 - cos^2x})(tan\frac{x}{2}) = \frac{1}{1+cosx}[/tex]
Step-by-step explanation:
Given
[tex](\frac{sin\ x}{1 - cos^2x})(tan\frac{x}{2})[/tex]
Required
Simplify
[tex](\frac{sin\ x}{1 - cos^2x})(tan\frac{x}{2})[/tex]
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Considering the denominator of the first bracket
[tex]1 - cos^2x[/tex]
This can be rewritten using different of two squares as follows:
[tex](1 - cosx)(1 + cosx)[/tex]
So, [tex]1 - cos^2x[/tex] can be replaced with [tex](1 - cosx)(1 + cosx)[/tex] in the given expression.
Also, from trigonometry identity
[tex]tan\frac{x}{2} = \frac{1 - cos x}{sin x}[/tex]
So, [tex]tan\frac{x}{2}[/tex] can be replaced with [tex]\frac{1 - cos x}{sin x}[/tex] in the given expression
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[tex](\frac{sin\ x}{1 - cos^2x})(tan\frac{x}{2})[/tex] becomes
[tex](\frac{sinx}{(1-cosx)(1+cosx)}) * \frac{1 - cos x}{sin x}[/tex]
[tex](\frac{sinx * (1 - cos x)}{(1-cosx)(1+cosx) * sin x})[/tex]
Cancel out similar expressions
[tex]\frac{1}{1+cosx}[/tex]
The expression cannot be further simplified.
Hence:
[tex](\frac{sin\ x}{1 - cos^2x})(tan\frac{x}{2}) = \frac{1}{1+cosx}[/tex]
The simplified form of the expression :
[tex]\frac{sinx}{1-cos^{2} x} .tan\frac{x}{2} =\frac{1}{1+cosx}[/tex]
We have some trigonometric formula as:
[tex]sin2x=2sinx= cosx\\2cos^{2} x=1+cos2x[/tex]
We have the expression
[tex]\frac{sinx}{1-cos^{2} x} .tan\frac{x}{2} \\=\frac{sinx}{sin^{2} x} .tan\frac{x}{2} \\=\frac{1}{sin x} .tan\frac{x}{2}\\=\frac{1}{2sin \frac{x}{2} cos\frac{x}{2}} .tan\frac{x}{2}\\=\frac{1}{2sin \frac{x}{2} cos\frac{x}{2}} .\frac{sin\frac{x}{2}}{cos\frac{x}{2}} \\=\frac{1}{2cos^{2} \frac{x}{2}}\\=\frac{1}{1+cos{x} }[/tex]
Therefore, the simplified form of the expression is:
[tex]\frac{sinx}{1-cos^{2} x} .tan\frac{x}{2} =\frac{1}{1+cosx}[/tex]
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