## Trigonometry (11th Edition) Clone

$$\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\cot x+\sec x\csc x$$ The expression is an identity.
$$\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\cot x+\sec x\csc x$$ 1) First, we examine the right side. $$A=\cot x+\sec x\csc x$$ The following identities would be applied now: $$\cot x=\frac{\cos x}{\sin x}\hspace{1cm}\sec x=\frac{1}{\cos x}\hspace{1cm}\csc x=\frac{1}{\sin x}$$ That means $A$ would be $$A=\frac{\cos x}{\sin x}+\frac{1}{\cos x}\frac{1}{\sin x}$$ $$A=\frac{\cos x}{\sin x}+\frac{1}{\sin x\cos x}$$ $$A=\frac{\cos^2 x+1}{\sin x\cos x}$$ 2) Now let's examine the left side. $$B=\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}$$ $$B=\frac{\tan x(1-\cos x)+\sin x(1+\cos x)}{(1-\cos x)(1+\cos x)}$$ $$B=\frac{\tan x-\tan x\cos x+\sin x+\sin x\cos x}{1-\cos^2 x}$$ Next we transform $\tan x=\frac{\sin x}{\cos x}$ and $\sin^2 x=1-\cos^2 x$ $$B=\frac{\frac{\sin x}{\cos x}-\frac{\sin x}{\cos x}\cos x+\sin x+\sin x\cos x}{\sin^2 x}$$ $$B=\frac{\frac{\sin x}{\cos x}-\sin x+\sin x+\sin x\cos x}{\sin^2 x}$$ $$B=\frac{\frac{\sin x}{\cos x}+\sin x\cos x}{\sin^2x}$$ $$B=\frac{\frac{\sin x+\sin x\cos^2 x}{\cos x}}{\sin^2x}$$ $$B=\frac{\sin x+\sin x\cos^2 x}{\sin^2x\cos x}$$ $$B=\frac{1+\cos^2 x}{\sin x\cos x}$$ (eliminate both numerator and denominator by $\sin x$) 3) We can see that $A=B$. The expression is therefore an identity.