([A=60quad B=B0],[hline(14)tan(AB)=(Can A-tan B)/(1+Tan A cos beta)]]

tan(A+B)=(tan A+tan B)/(1-tan A tan B)

If A=60⁰ and B=30⁰ , verify that tan (A-B)=(tan A-tan B)/(1+tan A tan B)

If A=60⁰ and B=30⁰ , verify that tan (A+B)=(tan A+tan B)/(1-tan A tan B)

If A=B=60o, verify that cos(A-B)=cos A cos B+sin A sin B(iii)sin(A-B)=sin A cos B-cos A sin B(iii)tan(A-B)=(tan A-tan B)/(1+tan A tan B)

For all values of A and Btan(A+B)=(tan A+tan B)/(1-tan A tan B) and tan(A-B)=(tan A-tan B)/(1+tan A tan B)

If A and B are acute angles such that tan A=(1)/(2),tan B=(1)/(3) and tan(A+B)=(tan A+tan B)/(1-tan A tan B), find A+B

if 3tan A tan B=1 then 2cos(A+B)

(tan A + tan B) / (tan (A + B)) + (tan A-tan B) / (tan (AB)) =

If A+B=(pi)/(2)rArr Tan A Tan B=1 , rArr Tan(A-B)=(tan A-tan B)/(1+tan A tan B)=(tan A-tan B)/(2) , rArr tan A=tan B+2tan(A-B) , (tan40^(@)+2tan10^(@))*(tan70^(@)-Tan20^(@))=

If A+B=(pi)/(2)rArr Tan A Tan B=1 , rArr Tan(A-B)=(tan A-tan B)/(1+tan A tan B)=(tan A-tan B)/(2) , rArr tan A=tan B+2tan(A-B) , x= Tan""(5 pi)/(28)+2Tan""(pi)/(7) Then x