How do you find the exact value of #sec(tan^-1(-3/5))#?

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Answer 1 · 2017-01-04T08:17:07

#sec(tan^(-1)(-3/5))=sqrt34/5#

Let #theta=tan^(-1)(-3/5)#

then #tantheta=-3/5# and

#sec^2theta=1+tan^2theta=1+9/25=34/25#

As #tantheta# is negative, as range of #tan^(-1)# is #-pi/2< theta < pi/2#, #theta# lies in Q4 and #sectheta# is positive and #sectheta=sqrt34/5#

and hence #sec(tan^(-1)(-3/5))#

= #sectheta#

= #sqrt34/5#