How do you simplify the expression #cosx+tanxsinx#?

Question

17602 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T01:59:19

#=cosx + sinx/cosx xx sinx#

#=cosx + sin^2x/cosx#

#=(cos^2x + sin^2x)/cosx#

#=1/cosx#

#=secx#

Hopefully this helps!

Answer 2 · 2016-09-02T02:01:03

#1/cos x = sec x#

#cos x + tan x sin x = cos x + (sinx/cos x) sin x#

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

#=((cos x)^2 +(sin x)^2)/cos x#

# = 1 / cos x = sec x#