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Prove that
`(sin theta - cos theta +1)/(sin theta + cos theta -1) = (1)/((sec theta - tan theta)). `

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LHS `=(sin theta - cos theta +1)/(sin theta + cos theta -1) `
` = ((sin theta)/(cos theta)-1 + (1)/(cos theta ))/((sin theta )/(cos theta) +1 - (1)/(cos theta)) `
[ on dividing num. and denom. by ` cos theta ` ]
` =(tan theta -1+ sec theta )/(tan theta +1 - sec theta )=((sec theta + tan theta -1))/((tan theta - sec theta +1)) `
` = ((sec theta + tan theta )-(sec^(2) theta - tan^(2) theta ))/((tan theta - sec theta +1)) " " [ because 1= sec^(2) theta - tan^(2) theta ] `
` =((sec theta + tan theta )[1- (sec theta - tan theta )])/((tan theta - sec theta +1)) `
` = ((sec theta + tan theta)(tan theta - sec theta +1))/((tan theta - sec theta +1))=(sec theta + tan theta). `
` RHS= (1)/((sec theta - tan theta)) `
` =(1)/((sec theta - tan theta))xx ((sec theta + tan theta ))/((sec theta + tan theta )) = ((sec theta + tan theta ))/((sec^(2)theta - tan^(2)theta))`
`= (sec theta + tan theta) " "[ because sec^(2)theta - tan^(2)theta =1].`
Hence,` LHS = RHS. `

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