Question

Verify the identity

tan (x +pi/2) = -cot x

Answer 1

I've gotten this far
tan (pi/2 +×)= sin (pi/2 +×)/ cos ( pi/2 +x)

Answer 2

HI!!

Answer 3

you need to use the "addition angle" formula for tangent

Answer 4

@misty1212 hey! that's quite a nostalgic formula

Answer 5

or use the addition angle formula for sine and cosine if you want to do what you wrote in the first step
either way

Answer 6

or you can,
\[\sin(\frac{\pi}{2}+x)=\cos(x)\]
\[\cos(\frac{\pi}{2}+x)=-\sin(x)\]

Answer 7

sine, cosine and tangent formulae
\[\sin(x+y)=sinxcosy+cosxsiny\]\[\cos(x+y)=cosxcosy-sinxsiny\]
\[\tan(x+y)=\frac{tanx+tany}{1-tanxtany}\]

Answer 8

OK that would give me
tan (pi/2 +×) = sin (×) cos (pi/2) + cos (×) sin (pi/2)
Right?

Answer 9

I'm confused @ Nishshant_Garg and @misty1212

Answer 10

@Nishant_Garg

Answer 11

yeah im here

Answer 12

that's not right allydiaz

Answer 13

\[\tan(\frac{\pi}{2}+x)=\frac{\sin(\frac{\pi}{2}+x)}{\cos(\frac{\pi}{2}+x)}\]
\[\frac{\sin(\frac{\pi}{2})cosx+\cos(\frac{\pi}{2}sinx)}{\cos(\frac{\pi}{2})cosx-\sin(\frac{\pi}{2})sinx}\]

Answer 14

sorry that sinx is not inside cos on the numerator

Answer 15

\[\sin(\frac{\pi}{2})=1\]\[\cos(\frac{\pi}{2})=0\]

Answer 16

OK so how would it cancel out to leave behind -cos/sin?

Answer 17

\[\frac{1 \times cosx+0 \times sinx}{0 \times cosx-1 \times sinx}=\frac{cosx}{-sinx}=-cotx\]

Answer 18

you replace sin pi/2 with 1 and cos pi/2 with 0

Answer 19

So cosx × sinx = cos x?