Question

sec^2(x)-1=csc^2(x)-cot^2(x)+sin^2(x)

Answer 1

so hmmm is that a question or a statement?

Answer 2

A question..

Answer 3

seems a bit occult though

what would the question be though?

Answer 4

I have to make these equal..

Answer 5

This is a problem in my math book

Answer 6

hmmm identity verification...h....k

Answer 7

\[\sec ^{2}x-1=\csc^2x-\cot^2x+\sin^2x\]

Answer 8

It looks better that way

Answer 9

hmmm

Answer 10

let us start by noticing something \(\bf \textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
1+tan^2(\theta)=sec^2(\theta)\implies tan^2(\theta)={\color{brown}{ sec^2(\theta)-1 }} \)

Answer 11

so we could say the left side is really just \(tan^2(x)\)

Answer 12

I noticed I wrote the wrong side down after rewriting everything. The real equation is \[1+\sin^2x=\csc^2x-\cot^2x+\sin^2x\]

Answer 13

hmmm

Answer 14

alright

let us poke the right-hand side then

Answer 15

since the right hand side is an identity can't we just cancel those?

Answer 16

hmmm actaully lemme fix that quick, I have the old version

Answer 17

\(\bf 1+sin^2(x)=csc^2(x)-cot^2(x)+sin^2(x)
\\ \quad \\ \quad \\

\\ \quad \\
csc^2(x)-cot^2(x)+sin^2(x)\implies \cfrac{1}{sin^2(x)}-\cfrac{cos^2(x)}{sin^2(x)}+\cfrac{sin^2(x)}{1}
\\ \quad \\
\cfrac{1-cos^2(x)+sin^2(x)sin^2(x)}{sin^2(x)}
\\ \quad \\
{\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta) }}\qquad thus
\\ \quad \\
\cfrac{1-cos^2(x)+sin^2(x)sin^2(x)}{sin^2(x)}\implies \cfrac{{\color{brown}{ sin^2(x)}}+sin^2(x)sin^2(x)}{sin^2(x)}
\\ \quad \\
\cfrac{\cancel{sin^2(x)}[1+sin^2(x)]}{\cancel{sin^2(x)}}\impliedby \textit{common factor}\)

Answer 18

but yes, you're correct

you can just use the \(\bf 1+cot^2(\theta)=csc^2(\theta)\implies 1=csc^2(\theta)-cot^2(\theta)\)

and not cancel out, but use the "1"

Answer 19

then it becomes \[1+\sin^2x\]

Answer 20

yeap

Answer 21

Another question!!

Answer 22

factoring \[\sin^3x-\cos^3x\]