Trig expression help?

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Answer 1 · 2018-06-23T18:19:17

The answer is #=-25/7#

Let #theta=tan^-1(4/3)#

Then,

#tantheta=4/3#

#tan^2theta+1=1/cos^2theta#

#1/cos^2theta=1+(4/3)^2=1+16/9=25/9#

#cos^2theta=9/25#

#costheta=3/5#

Therefore,

#cos2theta=2cos^2theta-1#

#=2*9/25-1#

#=18/25-1#

#=-7/25#

Finally,

#sec(2tan^-1(4/3))=sec2theta=1/cos(2theta)=1/(-7/25)=-25/7#