`(cos2B-cos2A)/(sin2B+sin2A)=tan(A-B)`

(cos2B-cos2A)/(sin2B+sin2A)=tan(AB)

Find the value of (cos2B-cos 2A)/(sin2A+sin2B) :

(cos 2B - cos 2 A)/( sin 2 A + sin 2 B) = tan (A -B).

Prove that (sin2A+sin2B)/(sin2A-sin2B)=tan(A+B)/tan(A-B)

If cos^(2)A-sin^(2)A=tan^(2)B, prove that 2cos^(2)B-1=cos^(2)B-sin^(2)B=tan^(2)A

Prove the following identities: tan^(2)A-tan^(2)B=(cos^(2)B-cos^(2)A)/(cos^(2)B cos^(2)A)=(sin^(2)A-sin^(2)B)/(cos^(2)A cos^(2)B)(sin A-sin B)/(cos A+cos B)+(cos A-cos B)/(sin A+sin B)=0

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

If A+B+C=90^(@) then (cos2A+cos2B+cos2C-1)/(sin A sin B sin C)=

Prove that ((sin 2A -sin 2B)/(cos2A + cos 2B))^(n) + ((cos 2A - Cos 2B)/(sin 2A + sin 2B))^(n) = 2 tan^(n) (A-B) or 0 According as n is even or n is odd.

Prove that (i) (sin^(2)A cos^(2)B - cos^(2)A sin^(2) B )=(sin^(2)A- sin^(2)B) (ii) (tan^(2)A sec^(2)B - sec^(2)A tan^(2)B)=(tan^(2)A- tan^(2)B)