If √ 3 tan 20-3 and 0° < 90°, find √3 ° the value 3 √3cos ③ + 2sin 6-6 tap²o 2
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Step-by-step explanation:
Given,
root(3)tan 2theta = 3
tan 2theta = 3/root(3) = root(3)
tan 2theta = tan 60
theta = 60/2 = 30
Therefore,
3root(3) cos theta + 2sin theta - 6 tan^2 theta
= 3root(3) cos 30 + 2sin 30 - 6 tan^2 30
= 3root(3)*root(3) + 2*1/root(2) - 6*(1/root(3))^2
= 3*3 + root(2) - 6/3
= 9 - 2 + root(2)
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