Question

(sinx)(cotx+cosxtanx)=cosx+sin^2x

Answer 1

\[\sin x(\cot x+\cos x\tan x)\]\[=\sin x\left(\frac{\cos x}{\sin x}+\cos x \frac{\sin x}{\cos x}\right)\] [Then cancellation and distribution.]

Answer 2

how do i cancel and distribute?

Answer 3

You can multiply \(\sin x\) to \(each\) of \(\frac{\cos x}{\sin x}\) and \(\cos x \frac{\sin x}{\cos x}\).

Answer 4

what happens with cos x?

Answer 5

cos x of the first expression remains, since only the sin x cancels, while the cos x of the second expression immediately cancels.

Answer 6

Like this:\[\sin x \left(\frac{\cos x}{\sin x}+\cos x \frac{\sin x}{\cos x} \right)\]\[=\sin x \frac{\cos x}{\sin x}+\sin x \cos x \frac{\sin x}{\cos x} \]\[=\cos x+\sin x\sin x\]

Answer 7

okay got it. thanks!