Step by step answer:
The identity should be:
[tex]\displaystyle \frac{(\sin x+ \cos x)^2}{\sin x \cos x} = 2 + \sec x\csc x[/tex]
You missed a + there in the numerator, otherwise it would not be an identity.
Expand the square on the numerator:
[tex]\displaystyle \frac{\sin^2x+ 2\sin x\cos x+\cos^2x}{\sin x \cos x} = 2 + \sec x\csc x[/tex]
Replace [tex] \sin^2x[/tex] with [tex]1-cos^2x\[/tex]:
[tex]\displaystyle \frac{1-\cos^2x+ 2\sin x\cos x+\cos^2x}{\sin x \cos x} = 2 + \sec x\csc x[/tex]
Combine like terms:
[tex]\displaystyle \frac{1+ 2\sin x\cos x}{\sin x \cos x} = 2 + \sec x\csc x[/tex]
Distribute the denominator through each term of numerator:
[tex]\displaystyle \frac{1}{\sin x \cos x }+ \frac{2\sin x\cos x}{\sin x \cos x } = 2 + \sec x\csc x[/tex]
Simplify the second fraction:
[tex]\displaystyle \frac{\sin x}{ \cos x }+ 2 = 2 + \sec x\csc x[/tex]
Use the reciprocal identities: sin(x)=1/csc(x) and cos(x) = 1/sec(x):
[tex]\sec x\csc x+ 2 = 2 + \sec x\csc x[/tex]
Re-ordering:
[tex]2+\sec x\csc x = 2 + \sec x\csc x[/tex]
