Question #a9ac6

Question

2134 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T19:01:33

see below

Use the following Properties:
1. Pythagorean Identity
$1+cot^2x=csc^2x$ and solve for $cot^2x$

Left Hand Side:
$cot^2x/(csc x-1)=(csc^2x-1)/(cscx-1)$

$=((csc x+1)(cscx-1))/(cscx -1)$ ->factor the difference of 2 squares above

$=((csc x+1)cancel((cscx-1)))/cancel((cscx -1))$

$=cscx+1$

$=1/sinx +1$

$=1/sinx+sinx/sinx$

$=(1+sinx)/sinx$

$:.=$ Right Hand Side