How do you simplify #sin(x) tan (x) + cos (x)#?

Question

37252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-26T16:58:41

#sin(x)tan(x)+cos(x)=sec(x)#

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

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

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

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

#=1/cos(x)#

#=sec(x)#

Answer 2 · 2016-06-13T07:02:12

#secx#

Given

#sinxtanx+cosx#

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

#=cosx(tanx*tanx+1)#
#=cosx(tan^2x+1)#
#=cosx*sec^2x #
#(" putting "(tan^2x+1)=sec^2x)#
#=secx" (putting" cosx*secx=1)#