# {(1/sinx) - (1/tanx)}^2 = (1-cosx)/(1+cosx). Prove LHS = RHS. Help anyone?

### 7 Answers

- Kali PrasadLv 68 years agoFavorite Answer
1/sin x - 1/tan x

= (1/ sin x - cos x/sin x)

= (1-cos x)/ sin x

so (1/sin x - 1/tan x)^2

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

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

= (1- cos x)^2/((1-cos x)(1+ cos x))

= (1- cos x)/(1+ cos x)

edit: for LHS to be defined sin x cannot be zero( if it is zero 1st term is undefined) , this takes care of tan x that is 2nd term ( sin x/ cos x) also . specifying just tan x and sin x defeats the purpose

and when sin ^2 x is in denominator then 1+ cos x can be cancelled

- Login to reply the answers

- silenderLv 44 years ago
above solutions are precise. yet another technique: Take a million+sinx =a, cosx=b. Then (a-b)^793d0c6d701ab1dda3c055e7572f1ea+b)^2 =? permit (a-b)^793d0c6d701ab1dda3c055e7572f1ea+b)^2=p (p+1793d0c6d701ab1dda3c055e7572f1ep-a million) = [(a-b)^2 + (a+b)^2]/[(a-b)^2 - (a+b)^2] =2(a^2+b^2793d0c6d701ab1dda3c055e7572f1e-4ab) = -a million/cosx ,putting fee of a,b. p=(a million-cosx793d0c6d701ab1dda3c055e7572f1e1+cosx)...

- Login to reply the answers

- big willie styleLv 68 years ago
working with the right side to turn it int the left side

rationalize

(1- cos x)^2/ sin ^2 x = having multiplied by conjugate of denominator (1 - cosx) and simplified

(1 - 2 cos x + cos ^2 x)/ sin ^2 x = multiplying out the square

csc ^2 x - 2 cot x csc x + cot ^2 x) = definition of sec is 1/ sinx defintion of cot = cos/ sin

(csc x - cot x)^2

[(1/sin x) - (1/tan x)]^2

as desired.

- Login to reply the answers

- How do you think about the answers? You can sign in to vote the answer.
- 8 years ago
Starting with:

{(1/sin x) - (1/tan x)}^2 =

We know that tan x = sin x / cos x, one of our basic trig identies, so LHS is equal to:

{(1/sin x) - (cos x /sin x)}^2 =

{(1- cos x)/ sin x}^2 =

{(1- cos x)^2 / (sin x)^2} =

It's time to break out another basic trig identity: (sin x)^2 + (cos x)^2 = 1 which can be rewritten as (sin x)^2 = 1 - (cos x)^2 so LHS is also equal to:

[(1 - cos x)^2 / {1 - (cos x)^2}] =

Remembering back to algebra, a^2 - b^2 = (a-b)(a+b) so LHS is equal to

[(1 - cos x)*(1-cos x)/ (1-cos x)(1+cos x)] =

[ (1 - cos x)/ (1 + cos x) ] QED, which is math-speak for Boo-yah, we did it!

- Login to reply the answers

- DouglasLv 78 years ago
Prove:

{1/sin(x) - 1/tan(x)}² = {1-cos(x)}/{1+cos(x)}

The common denominator for the left side is: sin(x)tan(x):

{[tan(x) - sin(x)]/[sin(x)tan(x)]}² = {1-cos(x)}/{1+cos(x)}

Multiply the left side by 1 in the form of cos(x)/cos(x)

{[cos(x)/cos(x)][tan(x) - sin(x)]/[sin(x)tan(x)]}² = {1-cos(x)}/{1+cos(x)}

We did the above step because this will turn any tan(x) into sin(x):

{[sin(x) - sin(x)cos(x)]/[sin²(x)]}² = {1-cos(x)}/{1+cos(x)}

The left side contains the factor sin(x)/sin(x) this equals 1 so it can be removed:

{[1 - cos(x)]/[sin(x)]}² = {1-cos(x)}/{1+cos(x)}

Move the squares to both the numerator and the denominator:

[1 - cos(x)]²/[sin(x)]² = {1-cos(x)}/{1+cos(x)}

Substitute 1 - cos²(x) for [sin(x)]²

[1 - cos(x)]²/[1 - cos²(x)] = {1-cos(x)}/{1+cos(x)}

1 - cos²(x) factors into {1 + cos(x)}{1 - cos(x)}:

[1 - cos(x)]²/[{1 + cos(x)}{1 - cos(x)}] = {1-cos(x)}/{1+cos(x)}

The left side contains a factor of [1 - cos(x)]/[1 - cos(x)] this can be removed, because it equals 1:

[1 - cos(x)]/[1 + cos(x)] = {1-cos(x)}/{1+cos(x)}

Q.E.D.

- Login to reply the answers

- 8 years ago
{(1/sinx)-(cotx)}^2

=[1/(sinx)^2]-[2(cotx)/sinx]+(cotx)^2

=[1-2cotxsinx+(cotx)^2(sinx)^2]/(sinx)^2

=[1-2cosx+(cosx)^2]/(1-(cosx)^2)

={(1-cosx)^2}/{(1-cosx)(1+cosx)}

=(1-cosx)/(1+cosx)

- Login to reply the answers