Question

(cosx)(tanx+sinx cotx)=sinx+cos^2x
Prove the identity

Answer 1

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

Answer 2

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

Answer 3

We know that \(\tan x = \dfrac{\sin x}{\cos x}\) and \(\cot x = \dfrac{\cos x}{\sin x}\)

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

Answer 4

\[\Huge\text{Q.E.D.}\]

Answer 5

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

Answer 6

http://en.wikipedia.org/wiki/Q.E.D.

Answer 7

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

Answer 8

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