# College Trig Question?

How do I verify the identity cotx/cscx - cscx/cotx = -sinxtanx ?

### 3 Answers

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- Wayne DeguManLv 74 weeks ago
cotx => cosx/sinx and cosecx = 1/sinx

so, cotx/cosecx = (cosx/sinx)/(1/sinx) = cosx

Hence, cosecx/cotx => 1/cosx

Then, cotx/cosecx - cosecx/cotx = cosx - (1/cosx)

so, RHS = (cos²x - 1)/cosx

Now, as sin²x + cos²x = 1, then cos²x - 1 = -sin²x

Then, (cos²x - 1)/cosx => -sin²x/cosx

i.e. -sinx(sinx/cosx) => -sinxtanx

:)>

- Iggy RockoLv 74 weeks ago
cotx/cscx - cscx/cotx =

cotx * sinx - cscx * tanx =(cosx/sinx) * sinx - (1/sinx) * sinx/cosx =

cosx - 1/cosx =

cos^2x/cosx - 1/cosx =

(cos^2x - 1)/cosx =

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

-sin^2x/cosx =

-sinx sinx/cosx =

-sinx tanx

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