Prove that cot a - tan a = 2 cos2a / sin2a

Question

Help!!! I"m so confused I've been writing in circles! 

Answer 1

cot(a)-tan(a)=2cos(2a)/sin(2a)

 

cos(a)/sin(a)-sin(a)/cos(a)=2cos(2a)/sin(2a)

combine the two terms on the left

 

[cos²(a)-sin²(a)]/sin(a)cos(a)=2cos(2a)/sin(2a)

you can continue working on the left side or work on the right side using double angle identities

[cos²(a)-sin²(a)]/sin(a)cos(a)=2[cos²(a)-sin²(a)]/2sin(a)cos(a)

factor out the two 2's on the right side and the two sides are identical

again, express cot and tan in terms of sin and cos and use double angle identities

 

Answer 2

When you have a problem like this, use a table of trig identities.  Here are the ones you need:

1. tan(a) = sin(a)/cos(a)
2. cot(a) = cos(a)/sin(a)
3. cos²(a) + sin²(a) = 1
4. sin(2a) = 2sin(a)cos(a)
5. cos(2a) = 1-2sin²(a)

 

Your problem is:

 

cot(a) - tan(a) = 2cos(2a)/sin(2a)

 

Let's deal with each side of the equation separately.  First the LHS:

 

cot(a) - tan(a) = cos(a)/sin(a) - sin(a)/cos(a) = [cos²(a) - sin²(a)]/sin(a)cos(a)    [Identities 1 and 2]

 

From identity 3, cos²(a) = 1 - sin²(a)

 

(1 - sin²(a) - sin²(a))/sin(a)cos(a)

 

= (1 - 2sin²(a))/sin(a)cos(a)

 

Now the RHS:

 

2cos(2a)/sin((2a) = 2(1 - 2sin^₂(a))/2sin(a)cos(a)

 

= (1 - 2sin²(a))/sin(a)cos(a)             [Identities 4 and 5]

 

RHS = LHS

 

Answer 3

cota-tana=2cos2a/sin2a

 

2cos(2a)/sin(2a)=2(cos²(a)-sin²⁽a))/(2sin(a)cos(a))=cos²(a)/sin(a)cos(a)-sin²a/sin(a)cos(a)=cos(a)/sin(a)-sin(a)/cos(a)=cot(a)-tan(a)