`cot ^(2) "" (pi)/(6) + cosec ""(5pi)/(6) + 3 tan ^(2) ""(pi)/(6) =6`

2 sin ^(2) "" (pi)/(6) + cosec ^(2) "" (7pi)/(6) cos ^(2) "" (pi)/(3) = 3/2

cos ^(2) (3 pi)/(5)+cos ^(2) (4 pi)/(5)=

tan^(-1)("tan"(2pi)/3)

Prove thet sin^(2) (pi//6) + cos^(2) (pi//3)- tan^(2) (pi//4) = (-1)/(2)

One of the values of ((cos (pi)/(6)-i sin (pi)/(6))^(1 / 2))((cos (pi)/(6)+i sin (pi)/(6))^(1 / 2)) is

Prove that 3 sin ""pi/6 sec ""pi/3 -4 sin ""(5pi)/(6) cot"" (pi)/(4) = 3 -4 xx 1/2 =1=R.H.S.

lim_(h rarr 0) ([2 sqrt(3) sin ((pi)/(6) + h) - cos ((pi)/(6) + h)])/(sqrt(3) h (sqrt(3) cos h - sin h)) is :

The maximum value of the function f(x) = sin (x+pi/6)+ cos (x+pi/6) in the interval (0, pi/2) occurs at