sec5pi/6=sec(pi-pi/6)=-sec(pi/6)=-2/sqrt3sec5π6=sec(π−π6)=−sec(π6)=−2√3
Now, sec^-1(-x)=pi-sec^-1x, |x|>=1sec−1(−x)=π−sec−1x,|x|≥1
Hence, sec^-1(sec5pi/6)=sec^-1(-2/sqrt3)=pi-sec^-1(2/sqrt3)=pi-pi/6=5pi/6.sec−1(sec5π6)=sec−1(−2√3)=π−sec−1(2√3)=π−π6=5π6.
Otherwise
If a fun. f :ArarrBf:A→B 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,y∈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