Question

(cosxtanx-tanx+2cosx-2)/tanx+2=cosx-1 can someone show me how to prove this?

Answer 1

Answer:

See detail below.

Step-by-step explanation:

A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:

(cosxtanx-tanx+2cosx-2)/(tanx+2)

Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.

Now to the actual problem: \we can make the following transformations on the left hand side:

$\frac{\cos x \tan x + 2\cos x -2}{\tan x +2}=\frac{\cos x \frac{\sin x}{\cos x} + 2\cos x -2}{\frac{\sin x +2\sin x}{\cos x} }=\\=\frac{\cos x \sin x - \sin x + 2\cos^2 x - 2\cos x}{\sin x + 2\cos x}=\\=\frac{\cos x \sin x +2\cos^2 x }{\sin x + 2\cos x}-1=\\=\cos x -1$

which is shown to be the same as the right hand side, which was to be shown.