How to prove (1+ cos 3x)/ (sin 3x) + (sin 3x)/ (1+ cos 3x) = 2 cosec 3x

LHS = (1+ cos 3x)/ (sin 3x) + (sin 3x)/ (1+ cos 3x)

Put over common denominator

LHS = [(1+cos3x)² + (sin3x)²]/(1+cos3x)(sin3x)

= [1 + cos² 3x + 2cos3x + sin² 3x]/(1+cos3x)(sin3x)

Since cos² 3x+ sin² 3x =1

LHS= (2 + 2cos3x)/ (1+cos3x)(sin3x)

= 2(1+cos3x) /(1+cos3x)(sin3x)

= 2/sin3x = 2cosec 3x = RHS

Work on the LHS,

(1+ cos 3x)/ (sin 3x) + (sin 3x)/ (1+ cos 3x)

Cross product them

[(1+cos 3x)^2+sin^2 3x]/[sin 3x(1+cos 3x)]

=[1+2cos 3x+cos^2 3x+ sin^2 3x]/[sin 3x(1+cos 3x)]

But cos^2 3x+sin^2 3x=1

=[2+2cos 3x]/[sin 3x(1+cos 3x)]

=2(1+cos 3x)/[sin 3x(1+cos 3x)]

1+cos 3x cancels out, we get

=2/sin 3x= csc 3x

(1+ cos 3x)/ (sin 3x) + (sin 3x)/ (1+ cos 3x) = 2 cosec 3x
{(1+cos3x)^2 + sin^2(3x)}/(1+cos3x).sin3x = 2cosec3x
=> 1+cos^2(3x)+2cos3x+sin^2(3x)/(1+cos3x).s… = 2cosec3x
=> 2(1+cos3x)/(1+cos3x)(sin3x) = 2cosec3x
=> 2/sin3x=2cosec3x
=> 2cosec3x=2cosec3x
Hence proved