Answer:
D) [tex]\frac{sin x}{1-cos x} X\frac{1+cos x}{1+cosx}[/tex]
Step-by-step explanation:
Given Expression
[tex]\frac{sin x}{1-cos x}[/tex]
Rationalizing
= [tex]\frac{sin x}{1-cos x} X\frac{1+cos x}{1+cosx}[/tex]
= [tex]\frac{sin x(1-cos x)}{1-cos^{2}x} }[/tex]
= [tex]\frac{sin x(1-cos x)}{sin^{2}x} }[/tex]
cancellation, sin x we get
[tex]= \frac{1-cos x}{sin x}\\ = \frac{2sin^{2} ({\frac{x}{2} } )}{2sin(\frac{x}{2} ) cos(\frac{x}{2}}[/tex] ( by using trigonometry formulas
[tex]1-cos A = 2 sin^{2} (\frac{A}{2} )[/tex]
[tex]sin A = 2 sin(\frac{A}{2} ) cos(\frac{A}{2} )[/tex] )
[tex]= \frac{sin({\frac{x}{2} } )}{ cos(\frac{x}{2}}[/tex]
[tex]= tan (\frac{x}{2} )[/tex]
Final answer:-
[tex]\frac{sin x}{1-cos x} = tan(\frac{x}{2} )[/tex]
