Question

Verify the identity. tan (x + pi/2) = -cot x
please show your work! :)

Answer 1

Do you know the sum rule for tangent?

Answer 2

i have it in front of me but im not sure what it means

Answer 3

ok... the sum rule is: \(\large tan(A+B)=\frac{tanA+tanB}{1-tanA \cdot tanB} \)
so let A = x, B = \(\frac{\pi}{2} \) and plug it in to the formula and simplify..

Answer 4

so would it be:
tan (x+π/2) = (tan x + tan π/2)/ (1-tan x)(tan π/2) @ByteMe

Answer 5

yes... but i'm thinking there's a problem here because tan(pi/2) is undefined...
It is better to rewrite it like this:
\(\large tan(A+B)=\frac{sin(A+B)}{cos(A+B)} \) =
that one ^^^ will work....

Answer 6

so that would b my final answer? @ByteMe

Answer 7

no... you're final answer is working to -cotx... you need to simplify it so that you end up with -cotx.

Answer 8

tan(x+π/2) = sin(x+π/2)/cos(x+π/2)

Answer 9

could u show me how?

Answer 10

yes... ^^^ now simplify that using sum formulas for sine and cosine..

Answer 11

sin(A+B) = sinAcosB + sinBcosA
cos(A+B) = cosAcosB - sinAsinB
again let A=x, and B=pi/2 and then simplify...

Answer 12

then how do i get to -cot x from there

Answer 13

sin(A+B) = sin(x + pi/2) = sinx cos(pi/2) + sin(pi/2) cos x = cos x

Answer 14

can you do the cosine part?

Answer 15

yeah one sec

Answer 16

cos (A+B) = cos(x+pi/2) = cos x cos(pi/2) - sin x sin(pi/2)

Answer 17

@ByteMe

Answer 18

that's correct... :)

Answer 19

notice that cos(pi/2) = 0, sin(pi/2) = 1
substitute those and simplify... wat u got????

Answer 20

sin x +cos x and
cos x - sin x

Answer 21

cos(x + pi/2) = cosx cos(pi/2) - sinx sin(pi/2) = cosx (0) - sinx (1) = 0 - sinx = -sinx

and since sin(x + pi/2) = cosx, then

tan(x + pi/2) = sin(x + pi/2) / cos(x + pi/2) = cosx / -sinx = - cosx/sinx = -cotx

Answer 22

k???

Answer 23

that makes sense. could you help me with one more plz?!?!

Answer 24

sure... just post it up as a separate question... :)

Answer 25

close this question and post a new question...

Answer 26

k