How do you show that #cosx/(1 - sinx) = secx+ tanx#?
1 Answer
Dec 4, 2016
We know that
#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
#cosx/(1 - sinx) xx (1 + sinx)/(1 + sinx) = (1 + sinx)/cosx#
#(cosx + cosxsinx)/(1 - sin^2x) = (1 + sinx)/cosx#
Use the identity
#(cosx(1 + sinx))/cos^2x = (1 + sinx)/cosx#
#(1 + sinx)/cosx = (1 + sinx)/cosx#
#LHS = RHS#
Identity Proved!
Hopefully this helps!