Answer 1

`(tan^2 60^@ + 4 cos^2 45^@ + 3 sec^2 30^@ + 5 cos^2 90)/(cosec 30^@ + sec 60^@ - cot^2 30^@)`   ....(i)

By trigonometric ratios we have

`tan 60^@ = sqrt3     cos 45^@ = 1/sqrt2    sec 30^@ = 2/sqrt3`

`cos 90^@ = 0      cosec  30^@ = 2      sec 60^@ = 2    cot 30^@ = sqrt3`

By substituting above values in (i), we get

`((sqrt3)^2 + 4.(1/sqrt3)^2 + 2 + [2/sqrt3]^2 + 5(0)^2)/(2 + 2sqrt2 (+ sqrt3)^2)`

`= (3 + 4. 1/2 + 3 4/3)/(4 - 3) = (3 + 2 + 4)/1 = 9`