cot A cot B=2,cos(A+B)=3/5rArr sin A.sin B=

cot A cot B=2,cos(A+B)=(3)/(5)rArr sin A sin B=

cot A+cot B+cot C+(cos(A+B+C))/(sin A sin B sin C)=

Cot A Cot B = 2, Cos (A+B) = 3/5 => sinA.sin B = (i) 2/5 (ii) 1/5 (iii) 4/5 (iv) 3/5

If Cot A Cot B = 2 & Cos (A+B) = 3/5 then Sin A Sin B

If A and B are complementary angles, prove that : cot A cot B - sin A cos B - cos A sin B = 0

Prove that : cot^(2) A - cot^(2) B = (cos^(2) A - cos^(2) B)/ (sin^(2) A sin^(2) B) = cosec^(2) A - cosec^(2)B

Prove that ((cos A + cos B)/( sin A - sin B )) ^(n) + ((sin A + sin B )/( cos A - cos B )) ^(n) ={{:(2 cot ^(n) ((A- B)/(2)),"," "if n is even."),(0,",""if n is odd."):}.

If f(theta)=det[[sin^(2)A,cot A,1sin^(2)B,cos B,1sin^(2)B,cos B,1sin^(2)C,cos C,1]], then (a) sin^(2)A+sin B+c(b)cot A cot B cot C(c)sin^(2)A+sin^(2)B+sin^(2)C(d)0