I was gone for a week and keep help with these. I think I'm sort of getting it, but I need help with the steps. It says "Transform the expression on the left to the one on the right."
There's 8 questions
1. (cscx)(tanx)(cosx)=1
2.(sin^2 x)(secx)(cscx)=tanx
3. (cos^2 x) (cscx)(secx)=cotx
4. tanx+cotx=(cscx)(secx)
5. sin x + cotx(cosx)+cscx
6. cscx-sinx= cotx(cosx)
7. secx-cosx=sinx(tanx)
8. tanx(sinx+cotx(cosx)=secx
You certainly don't have to explain them all, though that would be helpful. I just would really like some help setting up a few.
Thanks
There's 8 questions
1. (cscx)(tanx)(cosx)=1
2.(sin^2 x)(secx)(cscx)=tanx
3. (cos^2 x) (cscx)(secx)=cotx
4. tanx+cotx=(cscx)(secx)
5. sin x + cotx(cosx)+cscx
6. cscx-sinx= cotx(cosx)
7. secx-cosx=sinx(tanx)
8. tanx(sinx+cotx(cosx)=secx
You certainly don't have to explain them all, though that would be helpful. I just would really like some help setting up a few.
Thanks
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1. (1/sinx)(sinx/cosx)(cosx) which simplifies to 1
2. (sin^2 x)(1/cosx)(1/sinx)=sinx/cosx which is tanx
3. (cos^2 x)(1/sinx)(1/cosx)=(cosx/sinx) which is cotx
4. (sinx/cosx)+(cosx/sinx)=(sin^2 x + cos^2 x)/((sinx)(cosx))=1/((sinx)(cosx)) which is (cscx)(secx)
5. sinx+(cosx/sinx)(cosx)=sinx+(cos^2 x/sinx)=(sin^2 x + cos^2 x)/(sinx) which is cscx
6. (1/sinx)-(sinx)=(1-sin^2 x)/sinx=(cos^2 x)/(sinx)=(cosx/sinx)(cosx) which is (cotx)(cosx)
7. (1/cosx)-(cosx)=(1-cos^2 x)/(cosx)=(sin^2 x/cosx)=(sinx)(cosx/sinx) which is sinx(tanx)
8. I leave that for you to do on your own,as I've helped you through 7 similar questions.
2. (sin^2 x)(1/cosx)(1/sinx)=sinx/cosx which is tanx
3. (cos^2 x)(1/sinx)(1/cosx)=(cosx/sinx) which is cotx
4. (sinx/cosx)+(cosx/sinx)=(sin^2 x + cos^2 x)/((sinx)(cosx))=1/((sinx)(cosx)) which is (cscx)(secx)
5. sinx+(cosx/sinx)(cosx)=sinx+(cos^2 x/sinx)=(sin^2 x + cos^2 x)/(sinx) which is cscx
6. (1/sinx)-(sinx)=(1-sin^2 x)/sinx=(cos^2 x)/(sinx)=(cosx/sinx)(cosx) which is (cotx)(cosx)
7. (1/cosx)-(cosx)=(1-cos^2 x)/(cosx)=(sin^2 x/cosx)=(sinx)(cosx/sinx) which is sinx(tanx)
8. I leave that for you to do on your own,as I've helped you through 7 similar questions.