Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

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

Solution:

$\mathrm{LHS}=(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}$

$=\left[(\sec x \sec y)^{2}+(\tan x \tan y)^{2}-2(\sec x \sec y)(\tan x \tan y)\right]$

$-\left[(\sec x \tan y)^{2}+(\tan x \sec y)^{2}-2(\sec x \tan y)(\tan x \sec y)\right]$

$=\left[\begin{array}{llll}\sec ^{2} & x \sec ^{2} & y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y\end{array}\right]$

$-\left[\sec ^{2} x \tan ^{2} y+\tan ^{2} x \sec ^{2} y-2 \sec x \sec y \tan x \tan y\right]$

$=\sec ^{2} x \sec ^{2} y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y$

$-\sec ^{2} x \tan ^{2} y-\tan ^{2} x \sec ^{2} y+2 \sec x \sec y \tan x \tan y$

$=\sec ^{2} x \sec ^{2} y-\sec ^{2} x \tan ^{2} y+\tan ^{2} x \tan ^{2} y-\tan ^{2} x \sec ^{2} y$

$=\sec ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)+\tan ^{2} x\left(\tan ^{2} y-\sec ^{2} y\right)$

$=\sec ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)-\tan ^{2} x\left(\sec ^{2} y-\tan ^{2} y\right)$

$=\sec ^{2} x \times 1-\tan ^{2} x \times 1$

$=\sec ^{2} x-\tan ^{2} x$

$=1$

$=\mathrm{RHS}$

Hence proved.

Administrator

Leave a comment

Please enter comment.
Please enter your name.