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

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We have
` "LHS " = (sin theta - "cosec" theta)(cos theta- sec theta) `
` = (sin theta- (1)/(sin theta))(cos theta-(1)/(cos theta))=((sin^(2) theta-1))/(sin theta)* ((cos^(2) theta-1))/(cos theta) `
` =({-(1- sin^(2)theta)}*[-(1- cos^(2) theta)})/(sin theta cos theta ) `
` =((-cos^(2)theta)(-sin^(2) theta ))/(sin theta cos theta ) `
` [ because 1- sin^(2) theta= cos ^(2) theta " and " 1- cos^(2) theta= sin^(2)theta ] `
`=((cos^(2)theta sin^(2)theta))/(sin theta cos theta )= cos theta sin theta. `
` " RHS "=(1)/(tan theta +cot theta)`
` = (1)/((sin theta)/(cos theta)+(cos theta)/(sin theta))=(1)/((sin^(2)theta+cos^(2)theta)/(cos theta sin theta))= (cos theta sin theta)/((sin^(2)theta+cos^(2)theta))`
`=cos theta sin theta " "[because sin^(2)theta+cos^(2)theta=1]. `
` therefore LHS = RHS.`

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