`((1+tan33^(@))(1+tan12^(@)))/((1+tan18^(@))(1+tan27^(@)))=`

((1+tan13^(@))(1+tan32^(@)))/((1+tan12^(@))(1+tan33^(@))))=

The expression ((1+tan37^(@))(1+tan8^(@))(1+tan15^(@))(1+tan30^(@)))/((1+tan2^(@))(1+tan43^(@))) reduces to

If (tan46^(@)-tan1^(@)-1)(tan47^(@)-tan2^(@)-1)(tan48^(@)-tan30^(@)-1)...(tan89^(@)-tan44^(@)-1), then find the value of n.

Prove that tan12^(@)tan24^(@)tan48^(@)tan84^(@)=1

general solution of sec theta=(tan^(-1)(1)+tan^(-1)(2)+tan^(-1)(3))/(tan^(-1)1+tan^(-1)((1)/(2))+tan^(-1)((1)/(3))) is theta=

general solution of sec theta=(tan^(-1)(1)+tan^(-1)(2)+tan^(-1)(3))/(tan^(-1)1+tan^(-1)((1)/(2))+tan^(-1)((1)/(3))) is theta=

Solve tan^(-1)(x-1) +tan^(-1) x +tan^(-1)(x+1)= tan^(-1)(3x) .

Show that tan^(-1) (1/2) +tan^(-1) (1/8) = tan^(-1) (3/4) - tan^(-1)(1/18)