Verifying Identities II

nee

Aug 2008
36
0
tanx/(1 + cosx) + sinx/(1 - cosx) = cotx + secxcscx

HELPO!!!
 

Chris L T521

MHF Hall of Fame
May 2008
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2,046
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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...
 
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nee

Aug 2008
36
0
(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
(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 \(\displaystyle \cot x+\sec x\csc x\)...

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

Aug 2008
36
0
Oh ok, I simplified wrong. Thanks!!!