Chris L T521 MHF Hall of Fame May 2008 2,844 2,046 Chicago, IL May 10, 2010 #2 nee said: tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx HELPO!!! Click to expand... 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. Spoiler Apply the identity \(\displaystyle 1-\cos^2x=\sin^2x\) and simplify the result... Reactions: nee
nee said: tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx HELPO!!! Click to expand... 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. Spoiler Apply the identity \(\displaystyle 1-\cos^2x=\sin^2x\) and simplify the result...
N nee Aug 2008 36 0 May 10, 2010 #3 (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 ?
(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 May 2008 2,844 2,046 Chicago, IL May 10, 2010 #4 nee said: (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 ? Click to expand... It should simplify to \(\displaystyle \cot x+\sec x\csc x\)... \(\displaystyle \begin{aligned}\frac{\tan x-\sin x+\sin x+\sin x\cos x}{\sin^2x}&=\frac{\tan x}{\sin^2 x}+\frac{\sin x\cos x}{\sin^2 x}\\ &=\frac{1}{\cos x\sin x}+\frac{\cos x}{\sin x}\\ &=\ldots\end{aligned}\) Reactions: nee
nee said: (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 ? Click to expand... It should simplify to \(\displaystyle \cot x+\sec x\csc x\)... \(\displaystyle \begin{aligned}\frac{\tan x-\sin x+\sin x+\sin x\cos x}{\sin^2x}&=\frac{\tan x}{\sin^2 x}+\frac{\sin x\cos x}{\sin^2 x}\\ &=\frac{1}{\cos x\sin x}+\frac{\cos x}{\sin x}\\ &=\ldots\end{aligned}\)