How do you prove $sinx/(1-cosx) + (1-cosx)/sinx = 2csc x$ ?

Question

2663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-10T17:02:32

Please see below.

We take ,

$LHS=sinx/(1-cosx)+(1-cosx)/sinx$

$LHS=sin^2x/(sinx(1-cosx))+(1-cosx)/sinx$

$LHS$ = $(1-cos^2x)/(sinx(1-cosx))+(1-cosx)/sinxto[because sin^2x+cos^2x$ = $1]$

$LHS=((1-cosx)(1+cosx))/(sinx(1-cosx))+(1-cosx)/sinx$

$LHS=(1+cosx)/sinx+(1-cosx)/sinx$

$LHS=(1+cosx+1-cosx)/sinx$

$LHS=2/sinx$

$LHS=2cscx$

$LHS=RHS$

How do you prove sinx/(1-cosx) + (1-cosx)/sinx = 2csc xsinx/(1-cosx) + (1-cosx)/sinx = 2csc x?