answersLogoWhite

0


Best Answer

(tan x + cot x)/sec x . csc x

The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form.

Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x

The solution is:

[(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x)

[(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x)

[(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x)

then

sin x. sin x + cos x . cos x

sin2x+cos2x

=1

The answer is 1.

User Avatar

Wiki User

โˆ™ 2009-10-16 15:28:19
This answer is:
๐Ÿ™
0
๐Ÿคจ
0
๐Ÿ˜ฎ
0
User Avatar
Study guides

Math and Arithmetic

20 cards

What does h

What is the area of a cirle with a radius of 5 in

A yard is equal in length to three feet The function Fy takes a measurement in yards as input and returns a measurement in feet as output What input will return a value of 27

What is the height of a cone

โžก๏ธ
See all cards

Math and Arithmetic

21 cards

What are the laws of arithmetic

What property of real numbers as illustrated in 7 plus 2 equals 2 plus 7

Give example of closure property

What two multiples can be multiply by 621

โžก๏ธ
See all cards

Algebra

20 cards

Area of a triangle base of 10 and a height of 15

What is the maximum value of 4x 3y in the feasible region

What is the maximum value of 3x 4y in the feasible region

Shown here is the graph of the region described by which set of inequalities

โžก๏ธ
See all cards

Add your answer:

Earn +20 pts
Q: Solution for tan x plus cot x divided by sec x csc x?
Write your answer...
Submit
Related questions

Csc divided by cot squared equals tan multiplied by sec?

Yes.


What is the derivative of csc x?

The derivative of csc(x) is -cot(x)csc(x).


Can you simplify 1-cot x?

csc^2x+cot^2x=1


Csc squared divided by cot equals csc x sec. can someone make them equal?

cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)


How do you integrate the cscnx?

The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.


Cotangent squared plus one equals cosecant squared?

yes 1 + cot x^2 = csc x^2


Tan plus cot divided by tan equals csc squared?

(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.


How do you make 'cot' and 'csc' on a TI-84 graphing calculator?

From math class, some trigonometric identities: cot x = 1/tan x csc x = 1/sin x sec x = 1/cos x There are no built-in cot or csc formulas, so use the above. Remember that these give errors when tan x, sin x, or cos x are equal to 0.


Express csc theta in terms of cot theta theta is in quadrant 3?

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


What is the derivative of 3tanx-4cscx?

7


What is the anti derivative of cscxcotx?

∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C


How do you simplify sin theta times csc theta divided by tan theta?

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).


Verify that Cos theta cot theta plus sin theta equals csc theta?

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)


What is the answer of csc x plus cot x?

Without an "equals" sign somewhere, no question has been asked,so there's nothing there that needs an answer.Is it the sum that you're looking for ?csc(x) + cot(x) = 1/sin(x) + cos(x)/sin(x) = [1 + cos(x)] / sin(x)


How do you solve csc x sin x equals cos x cot x plus?

Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)


How do you simplify csc theta cot theta cos theta?

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


What is the second derivative of ln(tan(x))?

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


Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?

csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)


How do you simplify csc theta cot theta?

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.


What is co-tangent squared?

Start with the identity (sin a)2 + (cos a)2 = 1. Divide both sides by (sin a)2 to get1 + (cot a)2 = (csc a)2. Then subtract 1 from both sides. (cot a)2 = (csc a)2 - 1.


How do you simplify sec x cot x?

sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)


How do you simplify csc theta -cot theta cos theta?

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


What is cot x sin x simplified?

To simplify such expressions, it helps to express all trigonometric functions in terms of sines and cosines. That is, convert tan, cot, sec or csc to their equivalent in terms of sin and cos.


What is reciprocal in trigonometry?

The inverse of sine (sin) is cosecant (csc). The inverse of cosine (cos) is secant (sec). The inverse of tangent (tan) is cotangent (cot).


Are shifts sin cos tan equal to csc sec cot respectively?

No, they are the inverse functions, while csc, sec and cot are the reciprocal functions. To illustrate the difference, the inverse of f(x) = x+3 is f-1(x) = x-3 But the reciprocal of f(x) is 1/f(x) = 1/(x+3)