How do you evaluate #sec^-1(sec((5pi)/6))#?

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Answer 1 · 2016-07-14T16:22:14

Answer #=5pi/6#

#sec5pi/6=sec(pi-pi/6)=-sec(pi/6)=-2/sqrt3#

Now, #sec^-1(-x)=pi-sec^-1x, |x|>=1#

Hence, #sec^-1(sec5pi/6)=sec^-1(-2/sqrt3)=pi-sec^-1(2/sqrt3)=pi-pi/6=5pi/6.#

Otherwise

If a fun. #f :ArarrB# is an injection, i.e., one-to-one, then

#f(x)=f(y) rArrx=y, AAx,y in A.#

So, if #sec^-1(sec5pi/6)=theta, then, sectheta=sec5pi/6......(i)#

Now, #sec# is #1-1# in #[0,pi]-{pi/2}#, &, #5pi/6 in [0,pi]-{pi/2}#, from #(i)# we conclude that #theta....[=sec^-1(sec5pi/6)]...=5pi/6#