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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.

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Q: H ow do you verify that csc theta tan theta sec theta?

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

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

Ut is equual to tan(theta) / (sec(theta) + 1)

Tan^2

'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).

It also equals 13 12.

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

Yes.

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

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˜)

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

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.

Let m be the slope in percent and theta be the angle in question. tan (theta) = m/100 theta = arctan (m/100) To verify the result, we know the following: tan 0 = 0 tan (45 degrees) = 1 = 100% tan (90 degrees) = infinity For example, if 0 < m < 100%, then 0 < theta < 45 degrees.

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.

tan (theta x theta) : must square the value of the angle, theta, before applying the trig function, tangent.

Tan theta is a function of the number theta.

copy this and paste in your browsers address window http://www.wolframalpha.com/input/?i=tan+theta+%2B+sec+theta+%3D1

f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2

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

Below link may help.You need to memorize the basic formulas and it will be easy.pythagorean theoremc² = a² + b²Sine Theta (sin θ) = opposite/hypotenuse = a/cCosine Theta (cos θ) = adjacent/hypotenuse = b/cTangent Theta (tan θ) = opposite/adjacent = a/bCotangent Theta (cot θ) = adjacent/opposite = b/aSecant Theta (sec θ) = hypotenuse/adjacent = c/bCosecant Theta (csc θ) = hypotenuse/opposite = c/a

Sine Theta (sin θ) = opposite/hypotenuse = a/c Cosine Theta (cos θ) = adjacent/hypotenuse = b/c Tangent Theta (tan θ) = opposite/adjacent = a/b Cotangent Theta (cot θ) = adjacent/opposite = b/a Secant Theta (sec θ) = hypotenuse/adjacent = c/b Cosecant Theta (csc θ) = hypotenuse/opposite = c/a You may need to look on the link below for some sample calculations

It is 100*tan(theta).

1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1