verifying identities

[njacobson]

hello,

i need help verifying this identity: cotx + (sinx/1+cosx) = cscx

the first thing i did was to change the sinx to 1-cosx so i could cancel out the portion in parenthesis, but i am not sure if i can even do that.

after doing the previous step i changed the cotx to 1-cscx and am now left with 1-cscx = cscx and this is where i am stuck.

any help would be appreciated thank you!

 

[royhaas]

You need to use the correct identity, which is \(\displaystyle \sin^2(x) = 1 - \cos^2(x)\).

 

[tkhunny]

1) sin(x) / 1-cos(x) probably does not mean what you want. Add parentheses to clarify.

2) sin(x) = 1 - cos(x) ?! Where did you see that?

3) cot(x) = 1 - csc(x) ?! Where did you see that?

4) It is good that you are stuck. That would have been VERY magic if it would have worked.

 

[njacobson]

tkhunny:

one identity says that 1+cot2x=csc2x, therefore 1+cotx=cscx

or so my teacher says

if you have time, could you show me how you would solve it? thanks

 

[tkhunny]

There is no such identity. Forget you ever saw that. Pick an angle and prove it. \(\displaystyle \frac{\pi}{4}\) should do.

 

[Deleted member 4993]

tkhunny:

one identity says that 1+cot2x=csc2x,

you probably meant 1+cot[sup:3fhmjqri]2[/sup:3fhmjqri]x=csc[sup:3fhmjqri]2[/sup:3fhmjqri]x

therefore 1+cotx=cscx <<< What operation did that? - Not true except for very special points


or so my teacher says <<< Change teacher - FAST :twisted: :twisted:

if you have time, could you show me how you would solve it? thanks