Question

sinx/1-cosx+sinx/1+cosx=2 csc x
how do i prove that this is a identity?

Answer 1

\[sinx/1-cosx+sinx/1+cosx\]
\[sinx(1+cosx)+sinx*(1-\cos)/(1-cosx)(1+cosx)\]
\[sinx+sinxcosx+sinx-sinxcosx/1+cosx-cosx-cosx^2\]
\[2sinx/\sin^2x\]

Answer 2

\[2*sinx/sinx*sinx\]

Answer 3

1/sinx=cscx
2/sinx=2cscx

Answer 4

2*sinx/sinx*sinx cancel out sinx
so we are left with
2/sinx

Answer 5

LHS= sin x^2/(1-cos x) + sin x/(1+cos x)

=take LCM
we get,
{sin x(1+cos x)+sin x (1- cos x)} / (1-cos^2 x)

cancel +cos x and -cos x as it = 0

we are left with

{sin x (1+1)} / sin^2 x

=> 2 sin x/sin^2 x

cancel sin x from numerator and denominator

we get,
2 (1/sin x)


now
taking a look at RHS

2 cosec x

2(1/sinx) (as cosec is reciprocal of sin)

now LHS=RHS

hence proven ..

HOPE U CAN UNDERSTAND THE STEPS..I TRIED MY BEST!