Q: How can I solve tan of 3 theta -sqrt of 3 equals 0 for theta?

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If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)

sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = Â±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).

3tan (2t) = sqrt(3) tan (2t) = sqrt(3)/3 = 0.577 from tangent tables, 2t = 30 degrees t = 15 degrees

-2(cot2theta)

Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).

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If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)

tan theta = sqrt(2)/2 = 1/sqrt(2).

Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1

Yes, it is.

If tan theta equals 2, then the sides of the triangle could be -2, -1, and square root of 5 (I used the Pythagorean Theorem to get this). From this, sec theta is negative square root of 5. It is negative because theta is in the third quadrant, where cosine, secant, sine, and cosecant are all negative.

Yes. (Theta in radians, and then approximately, not exactly.)

It also equals 13 12.

Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.

tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)

It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.

sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = Â±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).

0.75