Show that :
`pi/3 - (5pi) /6 = -pi/2`

pi/4 + pi - pi/3 - pi/6

sin pi/3 = 2sin pi/6 cos pi/6

2*(pi)/(3)+(pi)/(6)=

sin^(-1)((-1)/2) (a) pi/3 (b) -pi/3 (c) pi/6 (d) -pi/6

Show that :(i) sin^(-1) [ sin. (3pi)/4] != (3pi)/4 (ii) tan^(-1) [ tan. (5pi)/6] != (5pi)/6 . What is its value ?

IF the lengths of the side of triangle are 3,5A N D7, then the largest angle of the triangle is pi/2 (b) (5pi)/6 (c) (2pi)/3 (d) (3pi)/4

If t a ntheta+s e ctheta=sqrt(3),\ 0

cot^2 pi/6 +cose c (5pi)/6 +3tan^2 pi/6=6

cos^-1 {-sin((5pi)/6)}= (A) - (5pi)/6 (B) (5pi)/6 (C) (2pi)/3 (D) -(2pi)/3

cot^(-1)(-sqrt3)= (a) -pi/6 (b) (5pi)/6 (c) pi/3 (d) (2pi)/3