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Question
Evaluate:
`2cos^2 60^0+3 sin^2 45^0 - 3 sin^2 30^0 + 2 cos^2 90 ^0`
Solution
On substituting the values of various T-ratios, we get:
`2cos^2 60^0+3 sin^2 45^0 - 3 sin^2 30^0 + 2 cos^2 90 ^0`
=`2xx(1/2)^2 + 3 xx(1/sqrt(2))^2 -3 xx (1/2)^2 + 2 xx (0)^2`
=`2xx1/4+3xx1/2-3xx1/4+0`
=`(1/2 +3/2-3/4)=((2+6-3)/4)=5/4`
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