Question

Please help!! Will medal!!
Verify the identity. cot(x- pi/2) = -tanx

Answer 1

@priyar @nincompoop @dramirez9 Please help!

Answer 2

do u know cot(pi/2 - x) =tanx ?

Answer 3

no does that mean the same thing?

Answer 4

not exactly the same thing..but u must have learnt this right?

Answer 5

I think I have seen it before

Answer 6

ok..and do u know cot(-x) = -cotx ?

Answer 7

yes, so what is the first step for this problem

Answer 8

these are the only things u need to solve this problem!

Answer 9

cot(pi/2 - x) ..take -1 common out from inside the bracket..what do u get?

Answer 10

@dschneider2016 ?

Answer 11

umm let me see, -cot(pi/2 - x) =tanx instead of cot(pi/2 - x)= -tanx ? @priyar

Answer 12

no..rewrite cot(pi/2 - x) after taking -1 common out

Answer 13

let-1 be just outside the brackets...

Answer 14

do you mean -cot(pi/2 - x)

Answer 15

no.i mean cot-(x - pi/2)

Answer 16

oh that makes sense. What is the next step @priyar

Answer 17

now.. this is in the form of cot( -y )..so what can this be written as (see abv.)

Answer 18

-cotx

Answer 19

is it "x" over there?

Answer 20

no so it would be -coty?

Answer 21

here what is our y?

Answer 22

so cot-(x - pi/2) can be written as -cot(x - pi/2) do u agree?

Answer 23

y= (x - pi/2)

Answer 24

yes I do

Answer 25

good

Answer 26

no.. i made a mistake ..

Answer 27

oh what as it?

Answer 28

so -cot(pi/2-x)=cot(x- pi/2) do u agree? i meant to ask this..

Answer 29

yes I do

Answer 30

did u understand how this came we have just taken the negative sign outside..

Answer 31

yes but is this all we have to do to verify the identity?

Answer 32

and we already know cot(pi/2 - x) = tanx put that in the abv expression..

Answer 33

just a nice reference...

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry.2C_shifts.2C_and_periodicity

Answer 34

i mean this exp : -cot(pi/2-x)=cot(x- pi/2)

Answer 35

Im kinda confused. If we were to combine everything to create a clear justification for verifing this identity what would it be (all the way to the answer)? @priyar @dumbcow

Answer 36

ok let me make things clearer..wait..

Answer 37

We know , cot(pi/2-x)=tanx
=> -cot(x-pi/2)=tanx
=> cot(x-pi/2) = -tanx

Answer 38

hence verified

Answer 39

so thats all I need?

Answer 40

yes..the explantion behind how these steps came is what i was making u understand..and u did pretty well..so i hope u understood..how to start when u see a problem like this and the 2 identities that we have used here..

Answer 41

yes I understand now! Thank you so much! you have been literally so helpful!

Answer 42

u r welcome! be sure to learn all the trig identities and remember them you will be using a lot of them frequently in maths...

Answer 43

yes! thanks so much! Im going to try to memorize them now

Answer 44

thats good!

Answer 45

\[ \cot(x- \pi/2) = -\tan x\]break down into component functions
\[\frac{\cos(x-\pi/2)}{\sin(x-\pi/2)} = -\frac{\sin x}{\cos x}\]but \(\sin x\) lags \(\cos x\) by \(\pi/2\) so \(\cos(x-\pi/2) = \sin x\)

\[\frac{\sin x}{\sin(x-\pi/2)} = -\frac{\sin x}{\cos x}\]

similarly, \(\sin(x-\pi/2) = -\cos x\)

so
\[\frac{\sin x}{-\cos x} = -\frac{\sin x}{\cos x}\]which is clearly true, so the identity is verified.