If A+B =90 then prove that √tanA+tanB. cotB/sinA. secB -sin squareB/cos cubeA=tanA

If A+B =90 then prove that √tanA+tanB. cotB/sinA. secB -sin squareB/

1.	If A+B =90 then prove that √tanA+tanB. cotB/sinA. secB -sin squareB/cos cubeA=tanA
Answer» Given,LHS=√tanAtanB+tanAcotB/sinAsecB-sin²B/cos²B=√tanAtan(90-A)+tanAcot(90-A)/sinAsec(90-A)-sin²(90-A)/cos²A=√tanAcotA+tanAtanA/sinAcosecA-cos²A/cos²A=√1+tan²A/1-1=√tan²A=tanA\xa0 U r new tenthian maths he.........u Paper ho gya ab dimag kharab mat kar