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

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`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.

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