Trying to prove a trig identity

[james_stewart]

Homework Statement

cos^2x-cotx
--------------- = cot^2x
sin^2x-tanx

Homework Equations

The Attempt at a Solution

every solution I get gives me a zero, not cot^2

 

[tiny-tim]

Welcome to PF!

Hi james_stewart! Welcome to PF!

(try using the X² tag just above the Reply box )

cos^2x-cotx
--------------- = cot^2x
sin^2x-tanx

Either write cot= cos/sin, tan = sin/cos,

or just multiply both sides by sin²x-tanx

 

[james_stewart]

i did and I'm not getting the proper results.

when i convert cot and tan to cos/sin and sin/cos i get

cos^2-cos^2
-------------
sin^2-sin^2

 

[tiny-tim]

(please use the X² tag just above the Reply box)

i did and I'm not getting the proper results.

when i convert cot and tan to cos/sin and sin/cos i get

cos^2-cos^2
-------------
sin^2-sin^2

No, you should get cos² - cos/sin on the top …

 

[james_stewart]

(please use the X² tag just above the Reply box)


No, you should get cos² - cos/sin on the top …

i did

and on the bottom i get sin²-sin/cos

 

[tiny-tim]

i did

and on the bottom i get sin²-sin/cos

ok, now put sin²-sin/cos as one fraction (ie with everything over the same denominator), and the same for cos²-cos/sin

 

[james_stewart]

ok, now put sin²-sin/cos as one fraction (ie with everything over the same denominator), and the same for cos²-cos/sin

That's how i did it. but where do i get cot² from this?

cos² - cos/sin
--------------
sin² - sin/cos

 

[tiny-tim]

erm … put sin² - sin/cos as one fraction (ie with everything over the same denominator), and the same for cos² - cos/sin

 

[rl.bhat]

That's how i did it. but where do i get cot² from this?

cos² - cos/sin
--------------
sin² - sin/cos

You can write numerator as
[cos²xsinx -cosx]/sinx. Then take cos(x) common.
Repeat the same thing for denominator and simplify.