Question

Prove the following identities (1-16)
${\left(\sec x\sec y+\tan x\tan y\right)}^2-{\left(\sec x\tan y+\tan x\sec y\right)}^2=1$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\left(\sec x\sec y+\tan x\tan y\right)}^2-{\left(\sec x\tan y+\tan x\sec y\right)}^2 \\ =\left[{\left(\sec x\sec y\right)}^2+{\left(\tan x\tan y\right)}^2-2\left(\sec x\sec y\right)\left(\tan x\tan y\right)\right] \\ -\left[{\left(\sec x\tan y\right)}^2+{\left(\tan x\sec y\right)}^2-2\left(\sec x\tan y\right)\left(\tan x\sec y\right)\right] \\ =\left[{\sec }^{2}x{\sec }^{2}y+{\tan }^{2}x{\tan }^{2}y-2\sec x\sec y\tan x\tan y\right] \\ -\left[{\sec }^{2}x{\tan }^{2}y+{\tan }^{2}x{\sec }^{2}y-2\sec x\sec y\tan x\tan y\right] \\ ={\sec }^{2}x{\sec }^{2}y+{\tan }^{2}x{\tan }^{2}y-2\sec x\sec y\tan x\tan y \\ -{\sec }^{2}x{\tan }^{2}y-{\tan }^{2}x{\sec }^{2}y+2\sec x\sec y\tan x\tan y \\ ={\sec }^{2}x{\sec }^{2}y-{\sec }^{2}x{\tan }^{2}y+{\tan }^{2}x{\tan }^{2}y-{\tan }^{2}x{\sec }^{2}y \\ ={\sec }^{2}x\left({\sec }^{2}y-{\tan }^{2}y\right)+{\tan }^{2}x\left({\tan }^{2}y-{\sec }^{2}y\right) \\ ={\sec }^{2}x\left({\sec }^{2}y-{\tan }^{2}y\right)-{\tan }^{2}x\left({\sec }^{2}y-{\tan }^{2}y\right) \\ ={\sec }^{2}x\times 1-{\tan }^{2}x\times 1 \\ ={\sec }^{2}x-{\tan }^{2}x \\ =1 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$