Question

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

Answer 1

Please help
i dont understand this

Answer 2

This is what i have but i dont think im doing it right
tan(x+pi/2)=sin(x+pi/2)/cos(x+pi/2)
sin(x)+sin(pi/2)/cos(x)+cos(pi/2)
sin(pi/2)=1
cos(pi/2)=0
sin(x)+1/cos(x)

Answer 3

tan(x+pi/2)=sin(x+pi/2)/cos(x+pi/2)
sin(x)+sin(pi/2)/cos(x)+cos(pi/2)
sin(pi/2)=1
cos(pi/2)=0
sin(x)+1/cos(x)
sin(x)cos(x)+cos(x)/cos^2(x)
cos^2(x)=1-sin^2(x)
1-sin^2(x)=(1+sin(x))(1-sin(x))
(cos(x))(sin(x)+1)=(1-sin(x))(1-sin(x))
cos(x)/1-sin(x)

^is this close?

Answer 4

\[\tan \left( x+\frac{ \pi }{ 2 } \right)=\frac{ \tan x+\tan \frac{ \pi }{ 2 } }{ 1-\tan x~\tan \frac{ \pi }{ 2 } },\]
divide the numerator and denominator by \[\tan \frac{ \pi }{ 2 }~on ~R.H.S\]
\[=\frac{ \frac{ \tan x }{ \tan \frac{ \pi }{ 2 } }+\frac{ \tan \frac{ \pi }{ 2 } }{ \tan \frac{ \pi }{ 2 } } }{ \frac{ 1 }{ \tan \frac{ \pi }{ 2 } }-\frac{ \tan x \tan \frac{ \pi }{ 2 } }{ \tan \frac{ \pi }{ 2 } } }\]
\[=\frac{ \tan x \cot \frac{ \pi }{ 2 } +1}{ \cot \frac{ \pi }{ 2 } -\tan x}\]
\[=\frac{ \tan x *0+1 }{0-\tan x }=\frac{ 1 }{ -\tan x }=-\cot x\]

Answer 5

THANK YOU SO MUCH!!!!!!!!!

Answer 6

yw