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Question
Evaluate the Following
`cot^2 30^@ - 2 cos^2 60^circ- 3/4 sec^2 45^@ - 4 sec^2 30^@`
Solution
`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`
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