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4tan^2u-1=tan^2u 3tan^2u=1 tan^2u=1/3 sq. root
4tan^2u-1=tan^2u
3tan^2u=1 tan^2u=1/3 sq. root of (tan^2u)=sq. root of (1/3) tanu=sq. root of (3)/3 tanu=sq. root of 1 Did I do
1 answer
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Tara
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3tan square á
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Solve: 3tan^2x-1=0. I got +/- (the sq. root of 3)/3 which is correct according to my study guide. However, I don't know how to
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solve the equation 3=4tan(2x+Pi/6) for x
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if cosA= -1/2 and π<A<3π/2 then find the value of 4tan^2A-3cosec^2A ?
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find y' of the functuion in terms of appropriate variable. simplify as far as possible.
y= 1+5^x^2-4tan(3x)
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