Q: What is Cos Theta minus Cos Theta?

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The identity for tan(theta) is sin(theta)/cos(theta).

The question contains an expression but not an equation. An expression cannot be solved.

The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant

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)

'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

Related questions

- cos theta

cosine (90- theta) = sine (theta)

Tan^2

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)

Cos theta squared

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

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

(Sin theta + cos theta)^n= sin n theta + cos n theta

The identity for tan(theta) is sin(theta)/cos(theta).

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

It is cotangent(theta).

The question contains an expression but not an equation. An expression cannot be solved.