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I know it has to be either tan x or cot x. My memory is not on my side today.
ok.. there are three trig identities. the most common is \[\cos^2x+\sin^2x=1\]
but there are two more equations.
\[1+\tan^2x=\sec^2x\] \[1+\cot^2x = \csc^2x \]
let's solve the right hand side of the equation to see who is the culprit lol xD so we expand \[(cscx-1)(cscx+1)\]
just use foil ... you will notice that O and I cancel out
Not familiar with FOIL (Sorry). My teacher calls it the "F word" (lol)
first outer inner last
\[(cscx-1)(cscx+1) = (cscx)(cscx)+(1)(cscx)+(-1)(cscx)+(1)(1)\]
fffffffff missed a sign on the last one should be +(-1)(1)
Oh ok! I think I see what you are getting at
\[(cscx-1)(cscx+1) = (cscx)(cscx)+(1)(cscx)+(-1)(cscx)+(-1)(1) = (cscx)^2+cscx-cscx-1\]
OY ! \[(cscx)^2+cscx-cscx-1 \]
the cscx-cscx cancels out
Then we can add the 1 right?
do you remember \[1+\cot^2x = \csc^2x ? \]
what do I need to do have \[\cot^2x \] by itself
subract the 1?
yes subtract on both sides..
Oh ok! So we just proved that cot^2 (x) is my answer right?
\[\cot^2x = \csc^2x-1\]
Would I just write it as cotx?
huh? you mean for (cotx)^2 = (csc^2x-1)
hmmm... if you don't forget this part \[\cot^2x = \csc^2x-1 \]
\[(cotx)^2 = (cscx)^2-1\] means the same
just don't write cot2x and csc2x-1 BIG NO NO!
Ok! Thank you for your time and teaching me!
if u observe the RHS, u see it results csc^2-1 there is also an identity saying that, cosec^2-cot^2=1 that implies cosec^2-1=cot^2...therefore theRHS is cot^2 therefore obviously LHS must be cot^2...therefore the answer is cot
^ done already lol
the term cosecant isn't used for the Pythagorean identities.
but there is one identity that is used beyond trig and that's the \[\cos^2x+\sin^2x = 1 \]