2cos2 B-1= tan2A ; cosA × cosB=?

Write tan²A as sin²A/cos²A .......

So, 2cos²B-1=sin²A/cos²A

Hence, 2cos²B x cos²A-cos²A=sin²A............

so, 2cos²B x cos²A=sin²A+cos²A...........

2(cos²Bxcos²A)=1

so your answer is +/-(1/sqrt2)...........

2cos²B-1=sin²A/cos²A

2cos²Bcos²A-cos²A=sin²A

2cos²Bcos²A=sin²A+cos²A

2cos²Bcos²A=1                       (where sin²A+cos²A=1)

cos²Acos²B=1/2.

thus, cosAxcosB=(+/-)root(1/2).