How do you verify # tan (x+pi/2)= -cot x#?

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Answer 1 · 2016-02-16T18:46:55

By applying the basic trigonometric relations. see below

head

Key-relation 1. #tanx=sinx/cosx#

Key-relation 2. #cotx=1/tanx=cosx/sinx#

Key-relation 3. #cos(a+b)=cosa*cosb -sina*sinb #

Key-relation 4. #sin(a+b)=cosa*sinb + sina*cosb #

Import results

Important result 1. #cos (pi/2)=0#
Important result 1. #sin (pi/2)=1#

Gathering

By using all the knowledge left so far and some mathematical tricks, we have:

#sin(x+(pi/2))=cosx*sin(pi/2) + sinx*cos(pi/2) #
#sin(x+(pi/2))=cosx#

Further:

#cos(x+(pi/2))=cosx*cos(pi/2) - sinx*sin(pi/2) #

#cos(x+(pi/2))=- sinx #

Finally:

#tan(x+(pi/2))=sin(x+(pi/2))/cos(x+(pi/2)) #

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

End of the proof!