Question

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

Answer 1

$$\text{ LHS }={(\sec x\sec y+\tan x\tan y)}^2-{(\sec x\tan y+\tan x\sec y)}^2$$
$$=[{(\sec x\sec y)}^2+{(\tan x\tan y)}^2-2(\sec x\sec y)(\tan x\tan y)]$$
$$-[{(\sec x\tan y)}^2+{(\tan x\sec y)}^2-2(\sec x\tan y)(\tan x\sec y)]$$
$$=[{\sec }^{2}x{\sec }^{2}y+{\tan }^{2}x{\tan }^{2}y-2\sec x\sec y\tan x\tan y]$$
$$-[{\sec }^{2}x{\tan }^{2}y+{\tan }^{2}x{\sec }^{2}y-2\sec x\sec y\tan x\tan y]$$
$$={\sec }^{2}x{\sec }^{2}y+{\tan }^{2}x{\tan }^{2}y-2\sec x\sec y\tan x\tan y$$
$$-{\sec }^{2}x{\tan }^{2}y-{\tan }^{2}x{\sec }^{2}y+2\sec x\sec y\tan x\tan y$$
$$={\sec }^{2}x{\sec }^{2}y-{\sec }^{2}x{\tan }^{2}y+{\tan }^{2}x{\tan }^{2}y-{\tan }^{2}x{\sec }^{2}y$$
$$={\sec }^{2}x({\sec }^{2}y-{\tan }^{2}y)+{\tan }^{2}x({\tan }^{2}y-{\sec }^{2}y)$$
$$={\sec }^{2}x({\sec }^{2}y-{\tan }^{2}y)-{\tan }^{2}x({\sec }^{2}y-{\tan }^{2}y)$$
$$={\sec }^{2}x\times 1-{\tan }^{2}x\times 1$$
$$={\sec }^{2}x-{\tan }^{2}x$$
$$=1$$
 = RHS
Hence proved.