Top Answer

If tan(theta) = x

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

= sqrt(1 + 1/x2)

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0'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

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

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).

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 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.

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 A = 1/2, then sin A = ? We use the Pythagorean identity 1 + cot2 A = csc2 A to find csc A, and then the reciprocal identity sin A = 1/csc A to find sin A. tan A = 1/2 (since tan A is positive, A is in the first or the third quadrant) cot A = 1/tan A = 1/(1/2) = 2 1 + cot2 A = csc2 A 1 + (2)2 = csc2 A 5 = csc2 A √5 = csc A (when A is in the first quadrant) 1/√5 = sin A √5/5 = sin A If A is in the third quadrant, then sin A = -√5/5.

No, they cannot all be negative and retain the same value for theta, as is shown with the four quadrants and their trigonemtric properties. For example, in the first quadrant (0

tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)

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

it is POSITIVE because tangent is said to be as OPPOSITE all over ADJACENT side of the triangle. since the opposite and adjacent sides of theta in Quadrant 3 are both negative, the quotient of two negative integers is POSITIVE. in third quadrant tanƟ= -O/-A

tan theta = sqrt(2)/2 = 1/sqrt(2).

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

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

tan(theta) = 1 then theta = tan-1(1) + n*pi where n is an integer = pi/4 + n*pi or pi*(1/4 + n) Within the given range, this gives theta = pi/4 and 5*pi/4

Tan theta is a function of the number theta.

First we must find what sign tan-theta will have. Sin-theta is negative, which means that theta must be in the 3rd or 4th quadrant of a graph. (from π radians (180˚) to 2π radians(360˚)) Now because cos-theta is greater than 0, it is positive, and cos-theta is only positive for values of theta in the 4th or 1st quadrant. The only possible choice is the 4th quadrant. Tan-theta equals sin-theta/cos-theta. Because sin is negative and cos is positive, tan-theta will be a negative value. Now to find the numerical answer. Sin-theta in a triangle is equal to the opposite/hypotenuse. In this problem, -0.5736 is the opposite. And the hypotenuse for all values of theta would be 1. Create a right triangle where the leg opposite to value theta has a value of 0.5736, and where the hypotenuse is 1. With a right triangle set up you can now use the Pythagorean theorem a2+b2=c2 Where a and b are values for the legs and c is the value for the hypotenuse. Using b for the leg opposite value theta, you manipulate the equation to work with the values found and you get: c2-b2=a2 Plug in the values for c and b now. 12-(-0.5736)2=a2 1-0.3290=a2 0.671=a2 square root both sides √0.671=a 0.819=a (all values are rounded off the the thousandths place) Now that you have the values for leg a and b, you can find the value for tan-theta. Tan-theta equals opposite/adjacent or b/a in this case. -0.5736/0.819=tan-theta -0.700≈tan-theta

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.

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

Suppose triangle ABC is right angled at C. Suppose you are given that the angle at B is theta. Thenif you know the length of AB (the hypotenuse), thenBC = AB*cos(theta) andAC = AB*sin(theta)if you know the length of BC, thenAB = BC/cos(theta) andAC = BC*tan(theta)if you know the length of AC, thenAB= AC/sin(theta) andBC = AC/tan(theta)

d/dx csc(x) = - csc(x) tan(x)

sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = Â±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).

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

The value of tan and sin is positive so you must search quadrant that tan and sin value is positive. The only quadrant fill that qualification is Quadrant 1.

Yes. (Theta in radians, and then approximately, not exactly.)

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