trig proofs

[Princezz3286]

tanxsinx/sec x -1=1+cos x
I am going to work with the left side....
((sinx/cosx)(sinx))/sec x-1
=((sin^2(x)/cos(x))/sec x-1
=??????? I'm not sure where to go from here.... there is probably an easier way to go about it but can someone please help?

 

[BigGlenntheHeavy]

\(\displaystyle \frac{tan(x)sin(x)}{sec(x)-1} \ = \ 1+cos(x)\)

\(\displaystyle \frac{sin^2(x)/cos(x)}{1/cos(x)-1} \ = \ \frac{sin^2(x)/cos(x)}{(1-cos(x))/cos(x)} \ = \ \frac{sin^2(x)}{1-cos(x)},\)

\(\displaystyle = \ \frac{1-cos^2(x)}{1-cos(x)} \ = \ \frac{(1+cos(x))(1-cos(x))}{1-cos(x)} \ = \ 1+cos(x).\)

\(\displaystyle Importance \ of \ Identities\)

\(\displaystyle Which \ would \ you \ rather \ evaluate?\)

\(\displaystyle \int\frac{tan(x)sin(x)}{sec(x)-1}dx \ or \ \int[1+cos(x)]dx.\)