Prove that #secx-cosx=tanxsinx#?

Question

6755 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-08T15:51:57

Please see below.

#secx-cosx#

= #1/cosx-cosx#

= #(1-cos^2x)/cosx#

= #sin^2x/cosx#

= #sinx/cosx xxsinx#

= #tanxsinx#

Answer 2 · 2016-10-08T15:55:19

I believe you have made an error in the question as:

#sec(x)-cos(x)-=1/cosx-cosx#

#:.sec(x)-cos(x) -=1/cosx-cosx*cosx/cosx#

#:.sec(x)-cos(x) -=1/cosx-cos^2x/cosx#

#:.sec(x)-cos(x) -=(1-cos^2x)/cosx#

#:.sec(x)-cos(x) -=(sin^2x)/cosx# (using #sin^2x+cos^2x-=1#)

#:.sec(x)-cos(x) -=sinx/cosx*sinx#

#:.sec(x)-cos(x) -=tanx*sinx#