Prove the following identity:
`(1+t a nalphat a nbeta)^2+(tanalpha-t a nbeta)^2=sec^2alphasec^2beta`

Explanation

`(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}=`
`\sec ^{2} \alpha \sec ^{2} \beta`
LHS `=(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}`
`=1+\tan ^{2} \alpha \tan ^{2} \beta+2 \tan \alpha \tan \beta+\tan ^{2} \alpha+`
`\tan ^{2} \beta-2 \tan \alpha \tan \beta`
`=1+\tan ^{2} \alpha \tan ^{2} \beta+\tan ^{2} \alpha+\tan ^{2} \beta`
`=\tan ^{2} \alpha(\tan ^{2} \beta+1)+1(1+\tan ^{2} \beta)`
`=(1+\tan ^{2} \alpha)(1+\tan ^{2} \beta)`
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