Question

Precalc:
Verity The Identity:
cot(x-(pi/2))=-tan(x)
no answer choices so i need to see how its done please

Answer 1

\(\bf cot\left(x-\cfrac{\pi}{2}\right) = -tan(x)\\
cot\left(x-\cfrac{\pi}{2}\right) \implies
\cfrac{cos\left(x-\cfrac{\pi}{2}\right)}{sin\left(x-\cfrac{\pi}{2}\right)}\\
\cfrac{cos(x)cos\left(\cfrac{\pi}{2}\right)+sin(x)sin\left(\cfrac{\pi}{2}\right)}
{sin(x)cos\left(\cfrac{\pi}{2}\right)-cos(x)sin\left(\cfrac{\pi}{2}\right)}\\
\textit{what's the sine of }\left(\cfrac{\pi}{2}\right)\ \ ?\\
\textit{what's the cosine of }\left(\cfrac{\pi}{2}\right)\ \ ?\)

Answer 2

cotx is an odd function. i.e cot(-x)=-cosx.Using this rule cot(pi/2-x)=tanx..we get cot(-(pi/2-x))=cot(x-pi/2)=-tanx

Answer 3

sorry..-cotx

Answer 4

@dlearner im sorry but i still dont understand. i get how we change -tan(x) to tan(-x) but you lose me after that. could you try to explain that part again

Answer 5

whats your doubt again??

Answer 6

how to prove cot(x-(pi/2))=tan(-x)

Answer 7

i mean in my reply :

Answer 8

u there @Ridict

Answer 9

yea im working it through and i get stuck at -cot(x)=-tan(x)

Answer 10

use the rule cot(90-x)=tanx. now multiply lhs and rhs by _.So we get rhs as -tanx which is result.now lhs is -cot(90-x).since cotx is odd fn,cot(-x)=-cotx.so change -cot(90-x)as cot(-(90-x)=cot(x-90).So now..both lhs and rhs are in req. form

Answer 11

ok i got it now thanx a ton