([22.],[-5in^(2)A sin^(2)B+cos^(2)A cos^(2)B=],[1)-1],[cdots r^(2)3cin^(2)A-cot^(2)B=1" then "sin A sin B]

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

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

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

Prove that sin ^(2) A cos ^(2) B+cos ^(2) A sin ^(2) B+cos ^(2) A cos ^(2) B+sin ^(2) A sin ^(2) B=1

(i) (1)/(sin^(2)a)-(1)/(sin^(2)B)=(cos^(2) a-cos^(2) B)/(sin^(2)a*sin^(2) B)

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

Prove that cos(A+B)cos(A-B)=cos^(2)A-sin^(2)B=cos^(2)B-sin^(2)A

Prove that: sin^(2)A=cos^(2)(A-B)+cos^(2)B-2cos(A-B)cos A cos B

2sin^(2)B+4cos(A+B)sin A sin B+cos2(A+B)=

In Delta ABC If,(sin^(2)A+sin^(2)B+sin^(2)C)/(cos^(2)A+cos^(2)B+cos^(2)C)=2 Then the triangle is