## rahrah Group Title (cosx)(tanx+sinx cotx)=sinx+cos^2x Prove the identity one year ago one year ago

1. RD⁴² Group Title

Are you sure it's not (cosx)(tanx+sinx cosx) = sinx + sinx cos²x ?

2. rahrah Group Title

oh crap, its actually (cosx)(tanx+sinx cotx)=sinx+cos^2x

3. RD⁴² Group Title

We know that $$\tan x = \dfrac{\sin x}{\cos x}$$ and $$\cot x = \dfrac{\cos x}{\sin x}$$ So $$(\cos x)(\tan x+\sin x \cot x) = (\cos x)\left(\dfrac{\sin x}{\cos x} + \sin x\cdot\dfrac{\cos x}{\sin x}\right)$$ $$=(\cos x)\left(\dfrac{\sin x}{\cos x} + \cos x\right)$$ $$= \boxed{\sin x + \cos^2 x}$$

4. RD⁴² Group Title

$\Huge\text{Q.E.D.}$

5. rahrah Group Title

Q.E.D. ? Thanks a lot for the help otherwise

6. RD⁴² Group Title
7. rahrah Group Title

wait, how'd you get sinx + cos^2 at the end? shouldn't sinx/cosx cross out the + cosx?

8. RD⁴² Group Title

No, because it's just addition. you can't cancel cos x out like that.|dw:1362452530886:dw|

9. rahrah Group Title

shouldnt that make the end result be tanx cosx + (cosx)^2? how did we get sin from the tan?

10. RD⁴² Group Title

$$\tan x = \dfrac{\sin x}{\cos x}$$ right? So $$\tan x\cos x\ = \dfrac{\sin x}{\cos x}\cdot \cos x = \sin x$$ And cos²x is just another way to write (cos x)²

11. rahrah Group Title

oh, i get it now. Thanks a lot for all your help!

12. RD⁴² Group Title