Question

Evaluate the following:

`cot^2 30^@ - 2 cos^2 60^circ- 3/4 sec^2 45^@ - 4 sec^2 30^@`

Answer 1

`cot^2 30^circ - 2 cos^2 60^circ - 3/4 sec^2 45^@ - 4 sec^2 30^circ`

By trigonometric ratios we have

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

By trigonometric ratios we have

`(sqrt3)^2 - 2[1/2]^2 - 3/4 (sqrt2)^2 - 4[2/sqrt3]^2`

`3 - 2 xx 1/4 - 3/4 xx 2 - 4 xx 4/3`

= `3/1 - 1/2 - 3/2 - 16/3`

= `(18 - 3 - 9 - 32)/6`

= `(18 - 44)/6`

= `(-26)/6`

= `(-13)/3`