1/sinx-sinx=cos/tan?

Question

can someone show me the steps to make this equation equal???? becasue i'm having major problems with identities!!!

Answer 1

The two identities that are important for this equation are:

(sin x)/(cos x) = tan x

(sin x)^2 + (cos x)^2 = 1

I found it easiest to work on the right side of the equation, and figure it in terms of sin x.

cos x / tan x
= cos x / (sin x / cos x)
= cos x * (cos x / sin x)
= (cos x)^2 / sin x

Next, the (cos x)^2 is equal to 1 - (sin x)^2, so substituting, we get:

= (1 - (sin x)^2) / sin x

Add finally, splitting the numerator,

= 1/sin x - (sin x)^2 / sin x
= 1/sin x - sin x

Which is the left side of the equation. This equation is therefore valid for all non-zero values of sin x.

Answer 2

If sinx isn't one million, then (one million + sinx) / (one million - sinx) = (one million + sinx)^2 / (one million - sin^2 x) = (one million + 2sinx + sin^2 x) / cos^2 x = sec^2 x + 2secx tanx + tan^2 x = 2sec^2 x + 2secx tanx - one million => Integration = 2tanx + 2secx - x + c. Edit: on the grounds that for f (x) = (one million + sinx) / (one million - sinx), sinx + one million could be 0, assuming this is not 0 will provide in basic terms a partly the final option answer. the perfect answer could be obtained with the aid of right here technique. (one million + sinx) / (one million - sinx) = [one million + cos(?/2 - x)] / [one million - cos(?/2 - x)] = 2cos^2 (?/4 - x/2) / 2sin^2 (?/4 - x/2) = cot^2 (?/4 - x/2) = coesc^2 (?/4 - x/2) - one million => fundamental = 2cot(?/4 - x/2) - x + c [notice the version between the two techniques. f(x) is defined for sinx + one million = 0, i.e., x = - ?/2 for which the 1st answer 2tanx + 2secx - x + c isn't defined while the 2d answer 2cot(?/4 - x/2) - x + c is defined.]

Answer 3

In one circle [0, 2π[
1/sinx
x ≠ π (180º), since 1/0 does not exist
With identities you have to work with either the left side or the right side, not both. Besides, you do not use =, because it is not an equation, it is an identity, and you just have to prove it. You use ⇒
1/sinx-sinx⇒
(common denominator (sinx))
(1 - sin^2x)/sinx⇒
since
sin^2x + cos^2x =1
then 1 - sin^2x = cos^2x
You substitute that in the identity
(1 - sin^2x)/sinx⇒ cos^2x/sinx
cos^2x/sinx can be also represented by
(cosx*cosx)/sinx

Now, you can do this
cos^2x/sinx ⇒ cosx*(cosx/sinx)

You know that tanx = sinx/cosx, so
tan^-1 x = cosx/sinx

You substitute that in the identity and you get to prove it
cosx*(cosx/sinx) ⇒ cosx*(1/tanx)⇒ cosx/tanx

And there you have it!!!!!!! I hope it helps

Answer 4

RHS
= cosx / tanx
= cosx / (sinx/cosx)
= (cosx)^2 / sinx
= 1-(sinx)^2 / sinx
= (1/sinx) - sinx
= LHS (Proven)

Answer 5

Ok. it is easy.
1/sinx-sinx = 1/sinx - sin(squared)x/sinx
=1-sin(squared)x/sinx
=cos(squared)/sinx
=cosx/sinx*cosx
sinx/cosx=tanx, so cosx/sinx=1/tanx
therefore, 1/tanx*cosx=cos/tan

get it?
things you need to know to understand this:
sin(squared)x+cos(squared)x=1
sinx/cosx=tanx

Answer 6

cosx/tanx
=cosx/(sinx/cosx)=cos^2x/sinx
=(1-sin^2x)/sinx
=(1/sinx)-sinx, as required

i hope that this helps

Answer 7

Start with sin^2(x) + cos^2(x) = 1
cos^2(x) = 1 - sin^2(x)
divide both sides by sin(x)
cos^2(x) / sin(x) = 1/sin(x) - sin(x)
cos(x) * cos(x)/sin(x) = 1/sin(x) - sin(x)
cos(x) / tan(x) = 1/sin(x) - sin(x)

{ sin^2(x) means sin(x) squared { sin(x) * sin(x) }, if you're not familiar with the symbols }

Answer 8

I assume you mean (1/sinx) - sinx = (cosx/tanx)?

Answer 9

1/s - s = c/t => t/s - ts = c => 1/c - ss/c = c => 1 - ss = cc => 1= ss + cc which is true. so the answer is yes.

Answer 10

Convert cos(x)/tan(x) into sines and cosines and they simplify:
tan(x) = sin(x)/cos(x) cos(x)/tan(x) = cos(x)/(sin(x)/cos(x)) = cos(x)*(cos(x)/sin(x) = (cos(x))^2/sin x

Now convert 1/sin(x) - sin(x) in to one fraction:
LCD is sin(x), sin(x)/1*(sin(x)/sin(x) = sinx^2/sin(x). 1/sin(x) - sin(x)^2/sin(x) = (1 -(sin(x))^2)/sin(x).

Now use the pythagorian Identity and subtract.
(sin(x))^2 + (cos(x))^2 = 1. 1 = (sin(x))^2 + (cos(x))^2. (1 -(sin(x))^2)/sin(x) = ((sin(x))^2 + (cos(x))^2 -(sin(x))^2). (sin(x))^2 -(sin(x))^2 + (cos(x))^2 )/sin(x). (cos(x))^2 )/sin(x)

Now compare.
(cos(x))^2 )/sin(x) = (cos(x))^2 )/sin(x)
Voila it is the same.
QED