Prove the following identities (1-16)
$(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}=1$
$\mathrm{LHS}=(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}$
$=\left[(\sec x \sec y)^{2}+(\tan x \tan y)^{2}-2(\sec x \sec y)(\tan x \tan y)\right]$
$-\left[(\sec x \tan y)^{2}+(\tan x \sec y)^{2}-2(\sec x \tan y)(\tan x \sec y)\right]$
$=\left[\begin{array}{llll}\sec ^{2} & x \sec ^{2} & y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y\end{array}\right]$
$-\left[\sec ^{2} x \tan ^{2} y+\tan ^{2} x \sec ^{2} y-2 \sec x \sec y \tan x \tan y\right]$
$=\sec ^{2} x \sec ^{2} y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y$
$-\sec ^{2} x \tan ^{2} y-\tan ^{2} x \sec ^{2} y+2 \sec x \sec y \tan x \tan y$
$=\sec ^{2} x \sec ^{2} y-\sec ^{2} x \tan ^{2} y+\tan ^{2} x \tan ^{2} y-\tan ^{2} x \sec ^{2} y$
$=\sec ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)+\tan ^{2} x\left(\tan ^{2} y-\sec ^{2} y\right)$
$=\sec ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)-\tan ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)$
$=\sec ^{2} x \times 1-\tan ^{2} x \times 1$
$=\sec ^{2} x-\tan ^{2} x$
$=1$
$=\mathrm{RHS}$
Hence proved.