Question

Prove that cot A-cot2A=cosec 2A

Answer 1

cot A= cos A/sin A rewrite is as

[(cos A)/(sin A)] - [(cos 2A)/(sin 2A)]

from the double angle identities we see that

sin 2A= 2(sin A)(cos A) now we can get the denominators equal so that we can write it as a single fraction consolidate the first term would be

[(cos A)/(sin A)]*[2cosA/2cos A]=[2(cos A)^2]/(2cosAsinA)

[2(cos A)^2 - cos2A] / [2cosAsinA]

double angle identity for cos 2A and now we can turn the denominator back to sin2A

[2(cos A)^2 - ( (cos A)^2 - (sin A)^2 )] / sin2A

[(cos A)^2 + (sin A)^2] / sin 2A

pythagorean identity

1 / sin 2A

reciprocal identity

csc 2A

QED