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+1
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https://prnt.sc/oj36ao

 

please explain each step.

i rearranged the right side into ln (1/cos^2 0) then I got stuck 

 Jul 23, 2019
 #1
avatar+105999 
0

Here is the question:

 

 Jul 24, 2019
 #2
avatar+105999 
+1

Interesting question :)

 

\(\frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1}{1+sin\theta}+\frac{1}{1-sin\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1-sin\theta +1+sin\theta}{1-sin^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{2}{cos^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \left(cos\theta \right)^{-2} \right]\\ \frac{2x+16}{x-1}=-2ln \left(cos\theta \right) \\ \frac{x+8}{x-1}=-ln \left(cos\theta \right) \\ \frac{x-1}{x-1}+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ 1+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ \frac{9}{x-1}=-ln \left(cos\theta \right) -1\\ \frac{x-1}{9}=\frac{1}{-ln \left(cos\theta \right) -1}\\ x-1=\frac{-9}{ln \left(cos\theta \right) +1}\\ x=\frac{-9}{ln \left(cos\theta \right) +1}+\frac{ln \left(cos\theta \right) +1}{ln \left(cos\theta \right) +1}\\ x=\frac{ln \left(cos\theta \right) -8}{ln \left(cos\theta \right) +1}\\\)

 

So it works     a=8 and b=1

 Jul 24, 2019

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