Question

I need to verify (sinx+tanx)/(cos+1)=tanx

Answer 1

work on the left-hand side (lhs):
\[lhs=\frac{\sin x+\tan x}{\cos x+1}\times \frac{\cos x-1}{\cos x-1}\]\[=\frac{(\sin x+\tan x)(\cos x-1)}{\cos^2x-1}\]from identity \(\sin^2x+\cos^2x=1,~\cos^2x-1=-\sin^2x\)\[=\frac{\sin x \cos x+\tan x \cos x-\sin x-\tan x}{-\sin^2x}\]simplify term-by-term\[=\frac{\sin x \cos x}{-\sin^2x}+\frac{\tan x \cos x}{-\sin^2x}-\frac{\sin x}{-\sin^2x}-\frac{\tan x}{-\sin^2x}\]\[=\frac{\cancel{\sin x}\cos x}{-\sin^{\cancel{2}}x}+\frac{\frac{\cancel{\sin x}}{\cancel{\cos x}}\cancel{\cos x}}{-\sin^{\cancel{2}}x}-\frac{\cancel{\sin x}}{-\sin^{\cancel{2}}x}-\frac{\frac{\cancel{\sin x}}{\cos x}}{-\sin^{\cancel{2}}x}\]let us clean it for a while\[=-\frac{\cos x}{\sin x}-\cancel{\frac{1}{\sin x}}+\cancel{\frac{1}{\sin x}}+\frac{1}{\sin x \cos x}\]\[=\frac{1}{\sin x \cos x}-\frac{\cos x}{\sin x}\]\[=\frac{1-\cos^2x}{\sin x \cos x}=\frac{\sin^2x}{\sin x \cos x}=\frac{\sin^\cancel{2}x}{\cancel{\sin x} \cos x}\]\[=\frac{\sin x}{\cos x}=\tan x\]this is equal to the right-hand side which is also \(\tan x\). Verified!

Answer 2

\[
\frac{\sin x+\tan x}{\cos x+1} = \frac{\sin x+\large \frac{\sin x}{\cos x}}{\cos x+1} =
\frac{\sin x*\cos x+\sin x}{\cos x (\cos x+1)} = \\
\frac{\sin x(\cos x+1)}{\cos x (\cos x+1)} = \frac{\sin x}{\cos x} = \tan x
\]