Question

Solve ${\left(1+\tan A\tan B\right)}^2+{\left(\tan A-\tan B\right)}^2={\sec }^{2}A{\sec }^{2}B$

Answer 1

$(1+tanAtanB{)}^2+(tanA-tanB{)}^2$

$=1+ta{n}^{2}Ata{n}^{2}B+2tanAtanB+ta{n}^{2}A+ta{n}^{2}B-2tanAtanB$

$=1+ta{n}^{2}Ata{n}^{2}B+ta{n}^{2}A+ta{n}^{2}B$

$=(1+ta{n}^{2}A)+ta{n}^{2}B(1+ta{n}^{2}A)$

$=(1+ta{n}^{2}A)(1+ta{n}^{2}B)$

$=se{c}^{2}A.se{c}^{2}B$