Respuesta :
Answer:
[tex]\frac{sinx}{1-cos x}[/tex] = [tex]cosecx[/tex] + [tex]cot x[/tex]
Step-by-step explanation:
To verify the identity:
sinx/1-cosx = cscx + cotx
we will follow the steps below;
We will take just the left-hand side and work it out to see if it is equal to the right-hand side
sinx/1-cosx
Multiply the numerator and denominator by 1 + cosx
That is;
[tex]\frac{sinx}{1-cos x}[/tex] = [tex]\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}[/tex]
open the parenthesis on the right-hand side of the equation at the numerator and the denominator
sinx(1+cosx) = sinx + sinx cosx
(1-cosx)(1+cosx) = 1 - cos²x
Hence
[tex]\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}[/tex] = [tex]\frac{sinx + sinx cosx}{1-cos^{2}x }[/tex]
But 1- cos²x = sin²x
Hence we will replace 1- cos²x by sin²x
[tex]\frac{sinx}{1-cos x}[/tex] = [tex]\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}[/tex] = [tex]\frac{sinx + sinx cosx}{1-cos^{2}x }[/tex] = [tex]\frac{sinx+sinxcosx}{sin^{2}x }[/tex]
=[tex]\frac{sinx}{sin^{2}x }[/tex] + [tex]\frac{sinxcosx}{sin^{2}x }[/tex]
=[tex]\frac{1}{sinx}[/tex] + [tex]\frac{cosx}{sinx}[/tex]
=[tex]cosecx[/tex] + [tex]cot x[/tex]
[tex]\frac{sinx}{1-cos x}[/tex] = [tex]cosecx[/tex] + [tex]cot x[/tex]
Note that;
[tex]\frac{1}{sinx}[/tex] = [tex]cosecx[/tex]
[tex]\frac{cosx}{sinx}[/tex] = [tex]cot x[/tex]
Answer:
Look at ttachment! Hope this helps!
Step-by-step explanation: