Question #cb109

1 Answer
Mar 1, 2017

We apply the identities #cottheta = costheta/sintheta# and #csctheta = 1/sintheta#, to obtain:

#(cosx/sinx)/(1/sinx - sinx) = secx#

#(cosx/sinx)/((1 - sin^2x)/sinx) = secx#

#cosx/sinx*sinx/(1 - sin^2x) = secx#

Now we can apply the identity #cos^2theta + sin^2theta = 1#:

#cosx/sinx * sinx/cos^2x = secx#

Eliminate:

#1/cosx = secx#

And this is true by the reciprocal identity #sectheta = 1/costheta#.

Our identity has been proved :).

Hopefully this helps!