Unsimplified Fraction

$let\quad \cfrac { \pi }{ 4 } -A=P,\cfrac { \pi }{ 4 } -B=Q\\ \\ \sin { \left( A+B \right) } =\cos { \left( P+Q \right) } \\ \\ \cos { \left( A-B \right) } =\cos { \left( Q-P \right) } \\ \\ \Rightarrow \cfrac { 1+5 }{ 1-5 } =\cfrac { \cos { \left( Q+P \right) } }{ \cos { \left( Q-P \right) } } \\ \\ \quad =\cfrac { \cos { P } \cos { Q } -\sin { P } \sin { Q } }{ \cos { P } \cos { Q } +\sin { P } \sin { Q } } \\ \\ \quad =\cfrac { 1-\tan { P } \tan { Q } }{ 1+\tan { P } \tan { Q } } =\cfrac { 1-X }{ 1+X } \\ \\ \Rightarrow X=-5$

Please,let me know is the answer is -5 or there is a mistake in my calculation.

Thanks.

easiest way is to simply put B=0 ..hence tanA=-1.5..then just expand second required expression ANS was -5

Simplifying the given equation (Sin A Cos B + Cos A Sin B)/(Cos A Cos B + Sin A Sin B) = ( Tan A + Tan B)/( 1 + Tan A Tan B) = -3/2

Hence Tan A + Tan B = (-3/2)(Tan A Tan B + 1)

Now Tan (Pi/4 - A) Tan (Pi/4 - B) = ((1 - Tan A)/(1 + Tan A) ((1 - Tan B)/(1 + Tan B))

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

= ( 1 + (3/2)( Tan A Tan B + 1) + Tan A Tan B)/( 1 - (3/2) (Tan A Tan B + 1) + Tan A Tan B)

= (5/2) ( 1 + Tan A Tan B)/ (-1/2) (1 + Tan A Tan B) = -5