Answer 1

`4/(cot^2 30^@) + 1/(sin^2 60^@) - cos^2 45^@` .....(i)

By trigonometric ratios we have

`cot 30^@ = sqrt3    sin 60^@ = sqrt3/2     cos 45^@ = 1/sqrt2`

By substituting above values in (i), we get

`4/(sqrt3)^2 + 1/(sqrt3/2)^2 - (1/sqrt2)^2`

`4/3 + 4/3 - 1/2 = 13/6`