Question

verify the identity cot^2x + csc^2x = 2csc^2x - 1
please help me and i dont just want the work i want to understand how to do it.

Answer 1

HI!!

Answer 2

rewrite everything in terms of sine and cosine, then do the arithmetic on the left, may work

Answer 3

can you walk me through it @misty1212

Answer 4

lets call cosine \(a\) and call sine \(b\)
then the left hand side is
\[\frac{a^2}{b^2}+\frac{1}{b^2}\]

Answer 5

add them up and you get
\[\frac{a^2+1}{b^2}\]

Answer 6

where are you getting cosine and sine

Answer 7

nvm this is easier than all that

Answer 8

\[\cot^2(x)=\csc^2(x)-1\] is all
we can prove that if you like

Answer 9

start with
\[\sin^2(x)+\cos^2(x)=1\] then divide all by \(\sin^2(x)\) to ge t
\[1+\cot^2(x)=\csc^2(x)\]

Answer 10

that means
\[\cot^2(x)=\csc^2(x)-1\]

Answer 11

plug that in and you get exactly what you want

Answer 12

\[\cot^2(x)+\csc^2(x)=\csc^2(x)+\csc^2(x)-1=2\csc^2(x)-1\] done

Answer 13

hey sorry @satellite73

Answer 14

\[\cot^2x+\csc^2=2\csc^2x-1\]
In order to prove this equality, I will just focus on the left side:
\[\cot^2x+\csc^2\]
And I will try to make them simple by using the equality:
\[\cot^2x=\csc^2x-1\]
So, then:
\[(\csc^2x-1)+\csc^2x\]
And now it's just a matter of simplying by operating the similar factors, I'll call csc^2x=u
just to make it simpler for you:
\[u-1+u\]
And then, operating:
\[2u-1\]
And, disarming the change of variable:
\[2\csc^2x-1\]
And that proves it.