Prove that: (a) `2sin^(2)(pi/2) + "cosec"^(2)((7pi)/2) cos^(2)(pi/3)=(3/2)^2` (b) `2sin^(2)((3pi)/4)+2cos^(2)(pi/4)+2sec^(2)(pi/3)=10`

Explanation

We have,
LHS `=2sin^(2)pi/6 + "cosec"^(2)(7pi)/6 cos^(2)pi/3`
`=2(sinpi/6)^(2)+{"cosec"(pi+pi/6)}^(2)(cospi/3)p^(2)`
`=2(1/2)^(2)+(-2)^(2) xx (1/2)^(2) = 1/2+1=3/2` =RHS [Since `"cosec"(pi + theta)=-"cosec"theta`]
Hence Proved.
(b) We have,
LHS `=2sin ^(2)(3pi)/4 + 2cos^(2)pi/4+2 sec^(2)pi/3`
`=2(sin(3pi)/4)^(2)+2(cospi/4)^(2) + 2(sec pi/3)^(2)`
`=2(sinpi/4)^(2)+2(cospi/4)^(2) + 2(sec pi/3)^(2)` `[therefore sin(3pi)/4 = sin(pi-pi/4)=sin pi/4]`
`=2(1/sqrt(2))^(2) + 2(1/sqrt(2))^(2) + 2(2)^(2)=1+1+8=10` = RHS. Hence proved.