Help me prove the following?

Question

Is it possible to prove the following:

[cotA]/[cotA + cosA] = [cosA]/[1 + sinA]

I get tan's everywhere and I just get lost trying to work it out.

Any help is appreciated. Thanks a lot.

Answer 1

Are you sure you copied correctly?  I got 1/[ 1 + sinA]  

[cotA]/[cotA + cosA] = [cosA]/[1 + sinA]

[cosA/sinA]/[(cosA/sinA) + cosA]

=[cosA/sinA]/[(cosA + cosAsinA)/sinA]

=[cosA/sinA]*[ sinA/(cosA + cosAsinA)]

=[cosA sinA ]/[sinA(cosA + cosAsinA)]  factor out cosA and reduce

=[cosA ]/[(cosA + cosAsinA)]  = 1/(1+sinA)

Answer 2

cos A/ sin A * sin A/cos A(1 + sin A)

1/1+sin A & not cos A/1 + sin A.

Answer 3

i hate trigonometry..

hehe..!

Answer 4

Hmm, this is a strange one, tbh.

Flip both sides of the identity over to give

[cot(A)+cos(A)]/[cot(A)] = [1+sin(A)]/[cos(A)]

Then you can split the left hand side into two fractions and simplify the first one to get

1 + [cos(A)]/[cot(A)] = [1+sin(A)]/[cos(A)]

Then if you use the idea that cot(A) = cos(A)/sin(A), you can simplify the left hand side to give

1 + sin(A) = [1+sin(A)]/[cos(A)]

Which shows that it only seems to work if cos(A) = 1, but most proofs have to be for all values of the variable, in this case A.

I'll take a closer look, and let you know if I get anywhere with it =]

Answer 5

Bro, I'm getting 1/(1 + sin A).

You're getting tans everywhere because you used the identity cot A = 1/tan A. You should've used cot A = cos A/sin A.

[cotA]/[cotA + cosA]

= ( [cos A]/[sin A] ) / ( [cos A/sin A + cos A] )

= [cos A] / [sin A . [cos A + cos A sin A] / [sin A] ] (Express the denominator in the first step as a single fraction)

= [cos A] / [cos A [1 + sin A] ] (The two sin A's cancel, factorise cos A + cos A sin A)

= 1 / [1 + sin A].

Hence, it's not possible to prove that [cotA]/[cotA + cosA] = [cosA]/[1 + sinA].

I hope that helps. :)

me07.

Answer 6

say what?? i don't know but i don't think so