Using Trigonometric Identities prove that [(tan^2x)(cosecx)]/sinx=sec^2x

You should begin by identifying all the Trigonometric Identities that may be useful in this problem. Specifically, cosecx=1/sinx tanx=sinx/cosx 1/cosx=secx and possibly tan^2x + 1= sec^2x. I began by changing cosecx into 1/sinx in hopes of simplifying the fraction: 

[Tan^2x(1/sinx)]/sinx 

I then simplified the fraction by multiplying the reciprocal of the top fraction (sinx/1) by the numerator and the denominator. This gave me:

tan^2x/sin^2x

I then substituted tan^2x in the numerator for the alternate sin^2x/cos^2x giving me:

(sin^2x/cos^2x)/sin^2x

Then I simplified the fraction multiplying by the reciprocal of the denominator (1/sin^2x) to both the numerator and the denominator of the fraction.

The denominator canceled out and both of the sin2^x cancel out in the numerator leaving me with 1/cos^2x which also equals sec^2x, completing the proof. 

1/cos^2x=sec^2x

sec^2x=sec^2x