Question

$(secAsecB+tanAtanB{)}^2-(secAtanB+tanAsecB{)}^2=1$

Answer 1

$LHS=(secAsecB+tanAtanB{)}^2-(secAtanB+tanAsecB{)}^2$

$=(secAsecB{)}^2+(tanAtanB{)}^2$ +2 sec A sec B tan A tan B

$=\left[\right(secAtanB{)}^2(tanAsecB{)}^2$+sec A tan B tan A sec B]

[Using $(a+b{)}^2=a^2+b^2+2ab]$

$=se{c}^{2}Ase{c}^{2}B+ta{n}^{2}Ata{n}^2$ B +2sec A sec B tan A tan B

$se{c}^{2}Ata{n}^{2}B-ta{n}^{2}Ase{c}^2$ B -2sec A sec B tan A tan B [Using $(ab{)}^2=a^2{b}^2]$

$=se{c}^{2}Ase{c}^{2}B-se{c}^{2}Ata{n}^{2}B+ta{n}^{2}Ata{n}^{2}B-ta{n}^{2}Ase{c}^{2}B.$

$=se{c}^{2}A(se{c}^{2}B-ta{n}^{2}B)+ta{n}^{2}A(ta{n}^{2}B-se{c}^{2}B)$

$=se{c}^{2}A1-ta{n}^{2}A1[\because se{c}^2\theta =1+ta{n}^2\theta \Rightarrow se{c}^2\theta -ta{n}^2\theta =1]$

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

=RHS Hence proved.