Can somebody prove this identity for me? cot A - tan A = 2cot 2A?

Question

Please show your working please. Thank you

Answer 1

cotA - tanA = 2cot(2A)

We're either going to choose the left hand side (LHS) or right hand side (RHS). I'm going to choose the RHS.

RHS = 2cot(2A)

Remember that cot(2A) = cos(2A)/sin(2A),so

RHS = 2[cos(2A) / sin(2A)]

By the double angle identites, cos(2x) = cos^2(x) - sin^2(x), and
sin(2x) = 2sinxcosx, so

RHS = 2 [ (cos^2(A) - sin^2(A)) / 2sinAcosA ]

We can cancel the 2 with the 2 in the denominator.

RHS = (cos^2(A) - sin^2(A)) / (sinAcosA))

And we can separate this into two fractions.

RHS = cos^2(A)/[sinAcosA] - sin^2(A)/[sinAcosA]

In each fraction, there is something to cancel.

RHS = cos(A)/sin(A) - sin(A)/cos(A)

And by definition, each of those is equal to

RHS = cot(A) - tan(A) = LHS

Answer 2

we start from RHS

2 cot 2A = 2 cos 2A / sin 2A

= 2 ( cos^2 A - sin ^2 A)/ sin 2 A

= 2 cos ^2 A / sin 2A - 2 sin ^2 A/ sin 2A

= 2 cos^2 A /(2 sin A cos A) - 2 sin ^2 A/( 2 sin A cos A )

= cos A/ sin A - sin A/cos A
= cot A - tan A
=LHS

Answer 3

Generally, you go from more complicated side first.

2 cot 2A = 2/ tan 2A = [1- (tan A)^2]/tan A = cot A - tan A

Reason: tan 2A = 2tan A/[1- (tan A)^2]

Answer 4

cosA/sinA-sinA/cosA=2cot2A
2(cos^2 A-sin^2 A)/2sinAcosA
by using identities
ie cos 2A=cos^A-sin^A
and sin2A=2*sinAcosA
2cos2A/sin2 A=2cot2A(hence proved)

Answer 5

just turn everything in terms of sines and cosines and then solve the identity.

Answer 6

LHS =1/tanA - tanA
=(1-(tanA)^2)/tanA
= 1/[tanA/(1-(tanA)^2)]
=2/[2tanA/(1-(tanA)^2)]
As 2tanA/(1-(tanA)^2)=tan2A
=2/tan2A
=2Cot2A