Posted by **Trig** on Sunday, November 9, 2008 at 2:16pm.

Prove that (3cos^2x + 8sinx-8/cos^2x) = (3sinx-5/sinx+1)

- Trigonometry -
**Reiny**, Sunday, November 9, 2008 at 3:03pm
your lack of the proper use of brackets make your equation much too ambiguous.

Furthermore, the brackets you do use, make no difference to the meaning.

If they are removed, there is no change in the order of operation, but ....

on the left side, who is divided by cos^2 x ?

is it -8/cos^2 x ? or

(8sinx-8)/cos^2 x or

(3cos^2x + 8sinx-8)/cos^2 x

each would give you a different result, the same is just as confusing on the right side.

- Trigonometry -
**Trig**, Sunday, November 9, 2008 at 3:31pm
(3cos^2x + 8sinx-8)/cos^2x = (3sinx-5)/sinx+1

- Trigonometry -
**Reiny**, Sunday, November 9, 2008 at 4:09pm
LS = (3(1-sin^x) + 8x - 8)/cos^2x

= (3-3sin^x+8x-8)/cos^2x

= -(3sin^2x - 8x + 5)/((1-sinx)(1+sinx))

= -(3sinx-5)(sinx-1)/((1-sinx)(1+sinx))

= (3sinx-5)(1 - sinx)/((1-sinx)(1+sinx))

= (3sinx-5)/(1+sinx)

= RS

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