Answer:
The answer is cscx(cscx - 1) ⇒ the second answer
Step-by-step explanation:
∵ csc²x = cot²x + 1
∴ [tex]\frac{cot^{2}x+1-1 }{1+sinx}=\frac{cot^{2}x }{1+sinx}[/tex]
Multiply the fraction by its conjugate 1 - sinx (up and down)
∴ [tex]\frac{cot^{2}x }{1+sinx}*\frac{1-sinx}{1-sinx}=\frac{cot^{2}x(1-sinx) }{1-sin^{2}x}[/tex]
∵ 1 - sin²x = cox²x
∴ [tex]\frac{cot^{2}x(1-sinx) }{cos^{2}x }[/tex]
∵ cot²x = cos²x/sin²x
∴ [tex]\frac{1-sinx}{sin^{x}}=\frac{1}{sin^{2}x}-\frac{sinx}{sin^{2}x}=csc^{2}x-cscx[/tex]
Take cscx as a common factor
∴ cscx(cscx - 1)
