Question

how do i reverse the signs when verifying that tan (x+pi/2)= -cotx

Answer 1

My approach could be wrong, but Im thinking this:

\[\tan(x+\frac{ \pi }{ 2 }) = \tan[-(-x-\frac{ \pi }{ 2 })]\] Because tangent is odd function, we can say:
\[\tan(-\theta) = -\tan \theta\]That takes care of the negative out front. Also, the period of tangent is pi, so -pi/2 would have the same value for tangent is pi/2. So rewriting -pi/2 as pi/2 is no problem:
\[-\tan(-x-\frac{ \pi }{ 2 }) \implies -\tan(\frac{ \pi }{ 2 }-x) = -cotx\]

Answer 2

thank you!!!!!

Answer 3

Np :3

Answer 4

Do you know All,silver,tea,cups formula...

Answer 5

no

Answer 6

If I get stuck, I go back to basics.
tan(x) = sin(x)/cos(x)

tan(a+b) = sin(a+b)/cos(a+b)
with b=90º
and the sum of angles formula
sin(a+90)= sin(a) cos(90) + cos(a) sin(90) which simplifies to cos(a)
cos(a+90)= cos(a) cos(90) - sin(a) sin(90) which simplifies to -sin(a)
and
cos(a)/-sin(a) = -cot(a)

Answer 7

Haha, I like that one.

Answer 8

Tan(pi/2+x) means it is in 2nd quadrent, so it becomes -ve

Answer 9

ohhhh