SinBtanB+sin(90°− B) how do I simplify this trig expression?

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Can someone please show me how to solve that? I am still learning how to get the hang of this. Thanks!

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Answer 1 · 2018-01-17T23:14:00

#1/cosB# or #secB#

We use the fact that #tanB=sinB/cosB# and that #sin(90-B)=cosB#:

#sinBtanB+sin(90-B)=sinB*sinB/cosB+cosB#
#=sin^2B/cosB+cos^2B/cosB#
#=(sin^2B+cos^2B)/cosB#

And now we use the fact that #sin^2B+cos^2B=1#

#=1/cosB=secB#