Question 568765
you change everything to sin and cos .. so what u do is you prove the left side since it will be easier

{{{(1/sinx + cosx/sinx)(1-cosx) = sinx}}}

since in the first term we have the same denominator they can be one fraction
so it will be 

{{{((1+cosx)/sinx)(1-cosx) = sinx}}}
now we just multiply the two terms
{{{((1+cosx)(1-cosx))/sinx = sinx}}}

now we know when we have two terms being multiplied and the sign is different we will have something like this

{{{(x-y)(x+y) = x^2-y^2}}}
same thing applies for the numerator in this equation
{{{(1+cosx)(1-cosx)/sinx = sinx}}}
{{{((1^2-cosx^2)/sinx) = sinx}}}
and we also know the trigonometric identity that says
{{{sinx^2 =1-cosx^2}}}

so we substitute that in for the numerator also and we get

{{{sinx^2/sinx=sinx}}}

dividing two same number with different exponents means subtracting the exponent so we will end up with

{{{sinx = sinx}}}


note : it is supposed to be the sin that is squared .. not the X .. but the website didnt allow me to do that so =/ ... but i hope this helps =)