If sin⁡Asin⁡B=32 and cos⁡Acos⁡B=520

From the given relation we have⇒ tan⁡A3=tan⁡B5=k( say ),( clearly k>0)Also 2sin⁡A=3sin⁡B.⇒2tan⁡A1+tan2⁡A=3tan⁡B1+tan2⁡B⇒23k1+3k2=3×5k1+5k2⇒41+5k2=51+3k2⇒k2=1/5⇒k=1/5so that tan⁡A=35,tan⁡B=1⇒tan⁡A+tan⁡B=3+55.