'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

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

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

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)

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.

csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4

csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.

With all due respect, you don't really want to know howto solve it.You just want the solution.csc(Î˜) = 1/sin(Î˜)tan(Î˜) = sin(Î˜)/cos(Î˜)csc(Î˜) x tan(Î˜) = 1/sin(Î˜) x sin(Î˜)/cos(Î˜) = 1/cos(Î˜) = sec(Î˜)

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)

There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.

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

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

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

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

Verify the identity:1/(cos Î¸)^2 - (tan Î¸)^2 = (cos Î¸)^2 + 1/(csc Î¸)^21/(cos Î¸)^2 - (sin Î¸)^2/(cos Î¸)^2 = (cos Î¸)^2 + (sin Î¸)^2 ?1 - (sin Î¸)^2/(cos Î¸)^2 = (cos Î¸)^2 + (sin Î¸)^2 ?(cos Î¸)^2/(cos Î¸)^2 = 1 ?1 = 1 TrueMethod 21/(cos Î¸)2 - (tan Î¸)2 =? (cos Î¸)2 + 1/(cscÎ¸)21/(cos Î¸)2-(sinÎ¸)2/(cos Î¸)2=? (cosÎ¸)2+ sin(Î¸)21/(cos Î¸)2[1-sin(Î¸)2]=? cos(Î¸)2+sin(Î¸)21/cos(Î¸)2(cos(Î¸)2)=? 11=1 True

It is -sqrt(1 + cot^2 theta)

The diagonal multiplied by sin(angle) gives one side of the rectangle and the diagonal times cos(theta) gives the other. So the area is (diagonal)2 x cos(theta) x sin(theta).

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

Zero. Anything minus itself is zero.

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

You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.

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

cosine (90- theta) = sine (theta)

Depending on your calculator, you should have an arcsin function, which appears as sin^-1. It's usually a 2nd function of the sin key. If you don't have this function, there are many free calculators you can download... just google scientific calculator downloads.Anyway, this inverse function will give you theta when you plug in the value of sin theta. Here's the algebra written out:sin(theta)=-0.0138arcsin(sin(theta))=arcsin(-0.0138)theta=.......The inverse function applied to both sides of the equation "cancels out" the sin function and yields the value of the angle that was originally plugged into the function, in this case theta. You can use this principle to solve for theta for any of the other trig functions:arccos(cos(theta))=thetaarctan(tan(theta))=thetaand so on, but calculators usually only have these three inverse functions, so if you encounter a problem using sec, csc, or cot, you need to rewrite it as cos, sin, or tan.sec=1/coscsc=1/sincot=1/tan

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