C ca.chick New member Joined May 7, 2007 Messages 19 May 31, 2007 #1 i am not sure how to figure this out: it has to be equal to each other and i got stuck: 1/tanx = cotx/ sin^2x + cos^2x i know that I should reduce the harder side, but i dont know how it would end up to be 1/tanx
i am not sure how to figure this out: it has to be equal to each other and i got stuck: 1/tanx = cotx/ sin^2x + cos^2x i know that I should reduce the harder side, but i dont know how it would end up to be 1/tanx
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 May 31, 2007 #2 This is just a matter of observation. You're making it too difficult. \(\displaystyle \L\\\frac{1}{tan(x)}=\frac{cot(x)}{\underbrace{sin^{2}(x)+cos^{2}(x)}_{\text{this is 1}}}\) I hope you know that 1/tan(x)=cot(x). See it?. You're done.
This is just a matter of observation. You're making it too difficult. \(\displaystyle \L\\\frac{1}{tan(x)}=\frac{cot(x)}{\underbrace{sin^{2}(x)+cos^{2}(x)}_{\text{this is 1}}}\) I hope you know that 1/tan(x)=cot(x). See it?. You're done.
