How to state the equivalent expression for cosxtanx?

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Answer 1 · 2017-11-19T02:09:28

#sin(x)#

One of our most important trig identities is:

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

Hence, we can rewrite the expression above as:

#cos(x)[sin(x)/cos(x)]#

Now, the cosines cancel out, so we're just left with:

#color(red)cancel(cos(x))[sin(x)/color(red)(cancel(cos(x)))]#

#=> sin(x)#

Hope that helped :)