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!
Write the right side as 1/(cscθ + cotθ) which can be rewritten as $$ \frac{1}{\frac{1}{sinθ} + \frac{cosθ}{sinθ}}$$ 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.
Did you misread the question? Multiplying them together gives [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]
Nope. My suggestion: (csc(x)-cot(x))(csc(x)+cot(x)) = csc^{2}(x)-cot^{2}(x) = csc^{2}(x)(1-cos^{2}(x)) = csc^{2}(x)sin^{2}(x) = 1. It seemed like the most straight-forward solution to me. Again no.