Prove the following identities: ( + )^2-( + )^2=1

Question

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

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Answer

$\text{L.H.S}=\big(\sec\text{x}\sec\text{x}+\tan\text{x}\tan\text{y}\big)^{2}-\big(\sec\text{x}\tan\text{y}+\tan\text{x}\sec\text{y}\big)^{2}$ $=\big(\sec\text{x}\sec\text{y}\big)^{2}+\big(\tan\text{x}\tan\text{y}\big)^{2}+2\sec\text{x}\sec\text{y}\tan\text{x}\tan\text{y}\\\ \ \ -\Big(\big(\sec\text{x}\tan\text{y}\big)^{2}+\big(\tan\text{x}\sec\text{y}\big)^{2}+2\sec\text{x}\tan\text{y}\tan\text{x}\sec\text{y}\Big)$ $\big[\text{using}\big(\text{a+b}\big)^{2}=\text{a}^{2}+\text{b}^{2}+2\text{ab}\big]$ $=\sec^{2}\text{x}\sec^{2}\text{y}+\tan^{2}\text{x}\tan^{2}\text{y}+2\sec\text{x}\sec\text{y}\tan{\text{x}}\tan{\text{y}}\\\ \ \ -\sec^{2}\text{x}\tan^{2}\text{y}-\tan^{2}\text{x}\sec^{2}\text{y}-2\sec\text{x}\sec\text{y}\tan\text{x}\tan\text{y}$ $\big[\text{using}\big(\text{ab}\big)^{2}=\text{a}^{2}\text{b}^{2}\big]$ $=\sec^{2}\text{x}\sec^{2}\text{y}-\sec^{2}\text{x}\tan^{2}\text{y}+\tan^{2}\text{x}\tan^{2}{\text{y}}-\tan^{2}\text{x}\sec^{2}\text{y}$ $=\sec^{2}\text{x}\big(\sec^{2}\text{y}-\tan^{2}\text{y}\big)+\tan^{2}\text{x}\big(\tan^{2}\text{y}-\sec^{2}\text{y}\big)$ $= \sec^{2}\text{x}1-\tan^{2}\text{x}1$ $\bigg[\because\sec^{2}\theta=1+\tan^{2}\theta\Rightarrow\sec^{2}\theta-\tan^{2}\theta=1\bigg]$ $=1+\tan^{2}\text{x}-\tan^{2}\text{x}$ $=1$ $=\text{R.H.S}$ $\text{Proved}$

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