# Verifying Identities II

#### nee

tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx

HELPO!!!

#### Chris L T521

MHF Hall of Fame
tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx

HELPO!!!
Move from LHS to RHS:

\displaystyle \begin{aligned}\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}&=\frac{\tan x(1-\cos x)}{1-\cos^2x}+\frac{\sin x(1+\cos x)}{1-\cos^2x}\\&=\frac{(\tan x-\sin x)+(\sin x +\sin x\cos x)}{1-\cos^2x}\end{aligned}

Can you finish this? If you can't, see the spoiler for the next hint.

Apply the identity $$\displaystyle 1-\cos^2x=\sin^2x$$ and simplify the result...

• nee

#### nee

(tanx - sinx) + (sinx + sinxcosx)
-------------------------------
sin^2x

= tanx + sinxcosx ?

How do you get the reciprocal identities at the end though. Shouldn't it simplify to cotx + secxcscx ?

#### Chris L T521

MHF Hall of Fame
(tanx - sinx) + (sinx + sinxcosx)
-------------------------------
sin^2x

= tanx + sinxcosx ?

How do you get the reciprocal identities at the end though. Shouldn't it simplify to cotx + secxcscx ?
It should simplify to $$\displaystyle \cot x+\sec x\csc x$$...

\displaystyle \begin{aligned}\frac{\tan x-\sin x+\sin x+\sin x\cos x}{\sin^2x}&=\frac{\tan x}{\sin^2 x}+\frac{\sin x\cos x}{\sin^2 x}\\ &=\frac{1}{\cos x\sin x}+\frac{\cos x}{\sin x}\\ &=\ldots\end{aligned}

• nee

#### nee

Oh ok, I simplified wrong. Thanks!!!