anonymous
  • anonymous
Verify the identity. cotx - pi / two. = -tan x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DanJS
  • DanJS
( )
Nnesha
  • Nnesha
can you take out the negative sign first and then use the identity
anonymous
  • anonymous
Im not sure how to do this at all

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More answers

Nnesha
  • Nnesha
well are you familiar with Cofunctions Identities ?
anonymous
  • anonymous
a little bit
Nnesha
  • Nnesha
remember cot(-x)=cot x not what would you get when you take out negative sign from (x-pi/2) ?
DanJS
  • DanJS
some things cot (pi/2 - x) = tan(x) cot(-x) = -cot x tan(-x) = -tan(x) . .
anonymous
  • anonymous
pi/2
Nnesha
  • Nnesha
\[\rm (x- \frac{\pi}{2})\] in other words -1 is a common factor so when we take out common factor we should divide both terms by the common factor so \[\cot(\frac{ x }{ -1} -\frac{ \frac{ \pi }{ 2 } }{ -1 })\] so when you divide x by negative one what would you get ? (pi/2) divide by -1 =?
anonymous
  • anonymous
- pi over 2
Nnesha
  • Nnesha
ugh nvm sorry it's (-pi/2) divide by -1 .
anonymous
  • anonymous
oh pi/2
anonymous
  • anonymous
?
Nnesha
  • Nnesha
..correct negative divide by negative = positive and negative divide by positive = negative xo x/-1 = -x so when you take out -1 from x u will get -x \[\rm \cot(-[\frac{ \pi }{ 2 }-x])\]and keep the common factor outside the bracket which is negative one now use the fact cot(-x)=-cot(x) (odd function ) and then use the identity
Nnesha
  • Nnesha
http://www.analyzemath.com/trigonometry/trigonometric_formulas.html here are trig identities u should copy these down n ur notebook :=))
anonymous
  • anonymous
What would i write as my final answer?
DanJS
  • DanJS
yeah, knowing the symmetries and coordinate sign changes on the circle helps alot to remember some things, symmetries
Nnesha
  • Nnesha
`verify` the identity you need to prove that L.H.S is equal to R.H.S
anonymous
  • anonymous
how?
Nnesha
  • Nnesha
what we just did ? we were working on left side to prove that it's equal to -tan(x)
Nnesha
  • Nnesha
look at the identity ^above (7th comment ) cot(pi/2-x)=tan(x) use that
Nnesha
  • Nnesha
let me ask u something if cot is an odd function then cot(-x)= -cot(x) here is an example \[\cot (\color{ReD}{-}4\pi) = -\cot (4\pi) \] so how would you write \[\cot (\color{ReD}{-}[\frac{\pi}{2}-x]) = ?\]

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