Prove the co function identity: cot(3.14/2-u) =Tan u

Question

3.14 is shown As pi

Answer 1

tan u = cot (π/2 - u)

tan u = cos (π/2 - u)  /  sin(π/2- u)

tan u = (cos π/2 cos u + sin π/2 sin u) / (sin π/2 cos u - cos π/2 sin u)

tan u = 0 + sin u / cos u - 0

tan u = sin u / cos u

tan u = tan u

 

In the strict form of an identity proof, only one side is transformed.

Use of the cofunction identities for sine and cosine would assume that they had been proven.

Answer 2

cot(pi/2 - u) = tan(pi/2 - u)

 

cos(pi/2 - u)          sin(u)

------------------   =  --------

sin(pi/2 - u)           cos(u)

 

Now cos(pi/2 - u) = sin(u) and sin(pi/2 - u) = cos(u):

 

sin(u)        sin(u)

--------   =  -------

cos(u)       cos(u)

 

1 = 1

 

QED

Answer 3

Often Trig expressions become manageable if you put the e? presision in terms of sin and cosine

 

cot(pi/2-u)

     =cos(pi/2-u)/sin(pi/2-u)

 

 I'm going to use the sin and cos difference angle identities 
sin(α – β) = sin(α)cos(β) – cos(α)sin(β)
cos(α – β) = cos(α)cos(β) + sin(α)sin(β)

Where α=pi/2

sin(pi/2 – β) = sin(pi/2)cos(β) – cos(pi/2)sin(β) 

cos(pi/2– β) = cos(pi/2)cos(β) + sin(pi/2)sin(β) 

 

Remember that

sin(pi/2)=1

cos(pi/2)=0

 

So for our case the identities simplify to

sin(pi/2 – β) = cos(β)  
cos(pi/2– β) = sin(β)

 

Plugging these into our expression

 

=cos(pi/2-u)/sin(pi/2-u)

=sin(u)/cos(u)=tan(u)

 

in this special case of α=pi/2

drawing the unit circle you should be able to see that 

sin(pi/2 – β) = cos(β)
cos(pi/2– β) = sin(β)

 

Without using the sin/cos of angle difference