answersLogoWhite

0


Best Answer

(1 + tanx)/sinx

Multiply by sinx/sinx

sinx + tanxsinx

Divide by sin2x (1/sin2x) = cscx

cscx + tan(x)csc(x)

tanx = sinx/cosx and cscx = 1/sinx

cscx + (sinx/cosx)(1/sinx)

sinx cancels out

cscx + 1/cosx

1/cosx = secx

cscx + secx

User Avatar

Wiki User

โˆ™ 2011-03-02 03:27:13
This answer is:
User Avatar
Study guides

Algebra

20 cards

A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

โžก๏ธ
See all cards
3.75
โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…
846 Reviews

Add your answer:

Earn +20 pts
Q: Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is the integral of cscx?

- ln (cscx + cotx) + C You use u substitution.


How do you rewrite y equals cscx to graph it?

You could try y = 1/sin(x) but I do not see how that helps.


Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?

pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!


Is 1 plus sinX divided by 1 plus cscX equal to sinX?

2


What is the derivative of cscx?

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


How do you verify the identity sinx cscx 1?

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.


How do you prove that the derivative of csc x is equals to -csc x cot x?

d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.


How do you find the derivative of - csc x - sin x?

d/dx (-cscx-sinx)=cscxcotx-cosx


What is the derivative of -cscxcotx?

Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.


What is the derivative of 1 divided by sinx?

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx


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

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.


What is the anti derivative of the square root of 1-x2?

-1

People also asked