Question #352ac

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Answer 1 · 2018-02-16T16:34:37

#tan (theta) #

#sec(theta) -= 1/cos(theta) #

#=> sin(theta)sec(theta) -= sin(theta)*1/cos(theta)#

#=sin(theta)/cos(theta)#

#=tan(theta)#

Answer 2 · 2018-02-16T16:37:56

#tan(theta)=sin(theta)sec(theta)#

#sec(theta)# is another way of writing #1/cos(theta)#

So #sin(theta)xxsec(theta) = sin(theta)xx1/cos(theta) = sin(theta)/cos(theta)#

but we also have #tan(theta)=sin(theta)/cos(theta)#

Thus #tan(theta)=sin(theta)sec(theta)#