Prove that 2 + sec(x) cosec(x) = (sin x + cos x)^2 / (sin x cos x).

2+secxcscx=(sinx+cosx)^2/(sinxcosx)
To prove, consider the left side of the equation.
2+secxcscx
Express the secant and cosecant in terms of cosine and sine, respectively.
=2+1/cosx*1/sinx
=2+1/(sinxcosx)
To add, express them as two fractions with same denominators.
=2*(sinxcosx)/(sinxcosx)+1/(sinxcosx)
=(2sinxcosx)/(sinxcosx) + 1/(sinxcosx)
=(2sinxcosx + 1)/(sinxcosx)
Apply the Pythagorean identity sin^2x+cos^2x=1 .
=(2sinxcosx+sin^2+cos^2x)/(sinxcosx)
=(sin^2x+2sinxcosx+cos^2x)/(sinxcosx)
And, factor the numerator.
= ((sinx +cosx)(sinx+cosx))/(sinxcosx)
=(sinx+cosx)^2/(sinxcosx)
Notice that this is the same expression that the right side of the equation have. Thus, this proves that the  2+secxcscx=(sinx+cosx)^2/(sinxcosx)  is an identity.