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Q: How do you simplify cos theta times csc theta divided by tan theta?

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

Cos theta squared

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

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

The side adjacent to theta divided by the hypotenuse, or the angle opposite of the right angle

- cos theta

Tan^2

cos2(theta) = 1 so cos(theta) = Â±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0

cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)

There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)

A - WORKWork = F.s cos (theta)

By converting everything to sines and cosines. Since tan x = sin x / cos x, in the cotangent, which is the reciprocal of the tangent: cot x = cos x / sin x. You can replace any other variable (like thetha) for the angle.

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