Question

"Verify the identity of (sinx cosx)^2/sinx cosx = 2 + secx cscx"

Answer 1

(sinx + cosx)^2/((sinx)(cosx)) = 2 + (secx)(cscx) 
(sinx + cosx)^2/((sinx)(cosx)) = 2 + 1/(sinxcosx); subtract 1/sinxcosx both sides 
(sinx + cosx)^2/((sinx)(cosx)) - 1/(sinxcosx)= 2; multiply through by sinxcosx 
(sinx + cosx)^2 -1 = 2(sinxcosx) 
sin^2 + 2sinxcosx + cos^2 - 1 = 2(sinxcosx); since sin^x + cos^2x = 1 
1 + 2sinxcosx -1 = 2sinxcosx 
2sinxcosx = 2sinxcosx

Answer 2

Step by step answer:

 

The identity should be:

$\displaystyle \frac{(\sin x+ \cos x)^2}{\sin x \cos x} = 2 + \sec x\csc x$

You missed a + there in the numerator, otherwise it would not be an identity.

Expand the square on the numerator:

$\displaystyle \frac{\sin^2x+ 2\sin x\cos x+\cos^2x}{\sin x \cos x} = 2 + \sec x\csc x$

Replace $\sin^2x$ with $1-cos^2x$:

$\displaystyle \frac{1-\cos^2x+ 2\sin x\cos x+\cos^2x}{\sin x \cos x} = 2 + \sec x\csc x$

Combine like terms:

$\displaystyle \frac{1+ 2\sin x\cos x}{\sin x \cos x} = 2 + \sec x\csc x$

Distribute the denominator through each term of numerator:

$\displaystyle \frac{1}{\sin x \cos x }+ \frac{2\sin x\cos x}{\sin x \cos x } = 2 + \sec x\csc x$

Simplify the second fraction:

$\displaystyle \frac{\sin x}{ \cos x }+ 2 = 2 + \sec x\csc x$

Use the reciprocal identities: sin(x)=1/csc(x) and cos(x) = 1/sec(x):

$\sec x\csc x+ 2 = 2 + \sec x\csc x$

Re-ordering:

$2+\sec x\csc x = 2 + \sec x\csc x$