How do you evaluate sec^-1(sec((5pi)/6))sec1(sec(5π6))?

1 Answer
Jul 14, 2016

Answer =5pi/6=5π6

Explanation:

sec5pi/6=sec(pi-pi/6)=-sec(pi/6)=-2/sqrt3sec5π6=sec(ππ6)=sec(π6)=23

Now, sec^-1(-x)=pi-sec^-1x, |x|>=1sec1(x)=πsec1x,|x|1

Hence, sec^-1(sec5pi/6)=sec^-1(-2/sqrt3)=pi-sec^-1(2/sqrt3)=pi-pi/6=5pi/6.sec1(sec5π6)=sec1(23)=πsec1(23)=ππ6=5π6.

Otherwise

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

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

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