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Q: How do you simplify csc theta tan 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)

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

-2(cot2theta)

They are theta = -34.99 degrees and 145.09 deg.

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta

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'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2

Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).

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

For a start, try converting everything to sines and cosines.

By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.

That depends on the value of the angle, theta. csc is short for "cosecans", and is the reciprocal of the sine. That is, csc theta = 1 / sin theta.

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

It is cotangent(theta).

cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)

If you want to simplify that, it usually helps to express all the trigonometric functions in terms of sines and cosines.

tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2

tan(x)*csc(x) = sec(x)

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