`sin^2 ((2pi)/3) + cos^2 ((5pi)/6) - tan^2 ((3pi)/2) `

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

tan^2 ((pi)/16) + tan^2 ((2pi)/16) + tan^2((3pi)/16).........+tan^2((7pi)/16 )'=35

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

cos^2(pi/8) +cos^2((3pi)/8) +cos^2((5pi)/8)+cos^2 ((7pi)/8)=2

Prove that the value of each the following determinants is zero: |sin^2(x+(3pi)/2)sin^2(x+(5pi)/2)sin^2(x+(7pi)/2)| |sin^.(x+(3pi)/2)sin^.(x+(5pi)/2)sin^.(x+(7pi)/2) | |sin^.(x-(3pi)/2)sin^.(x-(5pi)/2)sin^.(x-(7pi)/2)|

Prove that: sin ((8pi)/3) cos ((23pi)/6)+cos ((13pi)/3) sin ((35pi)/6)=1/2

Prove that: sin^2(pi/8)+sin^2((3pi)/8)+sin^2((5pi)/8)+sin^2((7pi)/8)=2

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

2sin^2(pi/6 )+c o s e c^2((7pi)/6)cos^2(pi/3)=3/2

Prove that: 2sin^2(pi/6)+cosec^2((7pi)/6)cos^2(pi/3)=3/2