(cot x / cos x) + (sec x / cot x) = csc x * sec^2 x

S

screambloodygore

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I've made a lot of progress since my last two posts yesterday- I did 15 questions out of the exercise and I solved all of them, except for this one, which is also supposed to be solved using only the 8 fundamental identities-- here's where I've gotten so far, and I'm pretty I'm getting close, but something's just not working for me no matter what:

(cot x / cos x) + (sec x / cot x) = csc x * sec^2 x

LHS: (cot x / cos x) + (sec x / cot x)

= ( (cot x * cot x) / (cos x * cot x) ) + ( (sec x * cos x) / (cos x * cot x) )

= (cot^2 x + sec x * cos x) / (cos x * cot x)

= (cot^2 x + 1) / (cos x * cot x)

= csc^2 x / (cos x * cot x)

= (csc^2 x / cos x) * (csc^2 x / cot x)

= ??

This is where I'm not sure what to do anymore.
 
After another 25 successful questions in a row:

This one is giving me a lot of trouble, too:

(cot x / (sec x - tan x) ) - (cos x / (sec x + tan x) ) = csc x + sin x

LHS:
= (cot x / (sec x - tan x) ) - (cos x / (sec x + tan x) )

= (cot x * (sec x + tan x) ) - (cos x * (sec x - tan x) ) / ( (sec x + tan x)*(sec x - tan x) )

= ( (cot x - cos x ) * (sec^2 x - tan^2 x) ) / (sec^2 x - tan^2 x)

= cot x - cos x

= (cos x / sin x) - cos x

= (cos x - (cos x * sin x) ) / sin x

...and I don't know where to go from here. If anyone could atleast steer me in the direction, that would probably help a lot- thanks.
 
I would suggest that you try using the fundamental identities, converting everything to sines and cosines first. See where that gets you. (You'd be amazed how often that helps.)

Eliz.
 
Re: After another 25 successful questions in a row:

Hello, screambloodygore!

\(\displaystyle \L\frac{\cot x}{\sec x\,-\,\tan x}\,-\,\frac{cos x}{\sec x\,+\,\tan x} \; =\;\csc x\,+\,\sin x\)
Click to expand...
Get a common denominator on the left side . . . something nice happens!

\(\displaystyle \L\;\;\;\frac{\cot x}{\sec x\,-\,\tan x}\,\cdot\,\frac{\sec x\,+\,\tan x}{\sec x\,+\,\tan x}\:-\:\frac{\cos x}{\sec x\,+\,\tan x}\,\cdot\,\frac{\sec x\,-\,\tan x}{\sec x\,-\,\tan x}\)

\(\displaystyle \L\;= \;\frac{\cot x(\sec x\,+\,\tan x)\,-\,\cos x(\sec x\,-\,\tan x)}{\sec^2x\,-\,\tan^2x}\)

But \(\displaystyle \sec^2x\,-\,\tan^2x\:=\:1\\)

\(\displaystyle \;\;\)so we have: \(\displaystyle \L\,\cot x\cdot\sec x\,+\,\cot x\cdot\tan x\,-\,\cos x\cdot\sec x\,+\,\cos x\cdot\tan x\)

But \(\displaystyle \,\cot x\cdot\tan x\,=\,1\,\) and \(\displaystyle \,\cos x\cdot\sec x\,=\,1\)

\(\displaystyle \;\;\)so we have: \(\displaystyle \L\,\sec x\cdot\cot x\,+\,1\,-\,1\,+\,\cos x\cdot\tan x\)

This becomes: \(\displaystyle \L\,\frac{1}{\cos x}\,\cdot\,\frac{\cos x}{\sin x}\:+\:\frac{\cos x}{1}\,\cdot\,\frac{\sin x}{\cos x} \;= \;\frac{1}{\sin x}\,+\,\sin x\;=\;\csc x\,+\,\sin x\)
 
Magnificient, thanks.
I just found that most of the work on identities I've been doing is completely optional, and I don't need to do anymore than the first couple sections of the chapter! hahaha!...

I'm in too deep now though.
 
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