
#1
Jan2014, 08:20 PM

P: 281

I am trying to prove that the equations below equal each other.
## (cscθ  cotθ) = (cscθ + cotθ)^{1} ## Besides letting the two sides equal each other, I have not been able to simplify each side independently to find the equivalent expression. I've had to remove terms from either side by multiplying each side simultaneously, but if anyone could show me how to do this proof one equation at a time, then that would be great! 



#2
Jan2014, 08:36 PM

P: 1,623

Just multiply them together and note that the result is 1. That is sufficient.




#3
Jan2114, 12:04 AM

Mentor
P: 21,008

$$ \frac{1}{\frac{1}{sinθ} + \frac{cosθ}{sinθ}}$$ A bit more work shows that the last expression above is equal to cscθ  cotθ. A quibble with your first sentence, above. One equation does not "equal" another. Equations can be equivalent (meaning they have exactly the same solution set), but never equal to one another. 



#4
Jan2114, 07:38 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

Proving equations[tex]\frac{csc(\theta) cot(\theta)}{csc(\theta)+ cot(\theta)}[/tex] which can be shown to be equal to 1 but is not obviously so. Did you miss the sign change? With problems of this kind I almost automatically change to "sine" and "cosine". [tex]csc(\theta) cot(\theta)= \frac{1}{sin(\theta} \frac{cos(\theta)}{sin(\theta)}[/tex] [tex]= \frac{1 cos(\theta)}{sin(\theta)}[/tex] Multiply both numerator and denominator by [itex]1+ cos(\theta)[/itex]: [tex]= \frac{1 cos^2(\theta)}{sin(\theta)(1+ cos(\theta))}[/tex] 



#5
Jan2214, 08:57 PM

P: 1,623




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