[" L "(4)" ) "quad " (n) "cos[cos((pi)/(2))+(pi)/(6)]],[" Thell.Fy "(1)/(2)+3" arc "tan(-1)+2arccos(-(1)/(2))]

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

If x in ((pi)/(6), (pi)/(3)) " then " int tan^(-1) sqrt((1 - cos 2x)/(1 + cos 2 x)) dx =

tan [(pi) / (4) + (1) / (2) (cos ^ (- 1)) (9) / (6)] + tan [(pi) / (4) - (1) / (2 ) (cos ^ (- 1)) (9) / (6)] =

tan((pi)/(4) +(1)/(2)"cos"^(-1)(a)/(b))+ tan((pi)/(4) - (1)/(2)"cos"^(-1)(a)/(b)) is :

tan((pi)/(4) +(1)/(2)"cos"^(-1)(a)/(b))+ tan((pi)/(4) - (1)/(2)"cos"^(-1)(a)/(b)) is :

Prove that (i) " tan"^(2) .(pi)/(3) + 2cos^(2) .(pi)/(4)+ 3 sec^(2).(pi)/(6)+ 4 cos^(2).(pi)/(2)=8 (ii) " sin ".(pi)/(6) " cos 0 + sin ".(pi)/(4) " cos " .(pi)(4) + " sin " .(pi)/(3) "cos " .(pi)/(6) =(7)/(4) (iii) " 4sin " (pi)/(6) " sin"^(2) (pi)/(3) + 3 " cos " .(pi)/(3) " tan ".(pi)/(4) = " cosec"^(2).(pi)/(2)=4

prove that tan ((pi) / (4) + (1) / (2) cos ^ (- 1) ((a) / (b))) + tan ((pi) / (4) - (1) / (2) cos ^ (- 1) ((a) / (b))) = (b) / (a) cos ^ (- 1) ((cos x + cos y) / (1 + cos x cos y) ) = 2tan ^ (- 1) ((tan x) / (2) (tan y) / (2))

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

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

"sin"^(2)(pi)/(6)+"cos"^(2)(pi)/(3)-"tan"^(2)(pi)/(4)+(1)/(2) is: