Question

( =frac{operatorname{los}(A-B)}{cos (A+B)}+frac{cos (C+D)}{cos (C-D)}=0 )
( Rightarrow frac{cos (A-B)}{cos (A+B)}=-frac{cos (C+D)}{cos (C-D)}=frac{-cos (C+D)+cos (C-D)}{-cos (C+D)-cos (C-D)} )
( = ) (A) ( frac{cos (A-B)-cos (A+B)}{cos (A-B)-sin C sin D}=2 operatorname{ses}(A+B) )
( Rightarrow frac{2 cos A cos B}{2 sin A sin B}=frac{1}{tan A tan B}=-tan C tan D )
( Rightarrow tan A tan B tan C tan D=-1 )

# If cos(A-B) cos(C+D) cos . then prove that tanA.tanB. tanC tanD = - 1.

Solution