How do you show that #cosx/(1 - sinx) = secx+ tanx#?

1 Answer
Dec 4, 2016

We know that #sectheta = 1/costheta# and #tantheta= sintheta/costheta#.

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

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

Multiply the left side by the conjugate of the denominator. The conjugate of #a + b# is #a- b#, for example.

#cosx/(1 - sinx) xx (1 + sinx)/(1 + sinx) = (1 + sinx)/cosx#

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

Use the identity #sin^2theta + cos^2theta= 1-> cos^2theta = 1- sin^2theta#.

#(cosx(1 + sinx))/cos^2x = (1 + sinx)/cosx#

#(1 + sinx)/cosx = (1 + sinx)/cosx#

#LHS = RHS#

Identity Proved!

Hopefully this helps!