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प्रश्न
`(cos^3 theta +sin^3 theta)/(cos theta + sin theta) + (cos ^3 theta - sin^3 theta)/(cos theta - sin theta) = 2`
उत्तर
LHS= `(cos^3 theta +sin^3 theta)/(cos theta + sin theta) + (cos ^3 theta - sin^3 theta)/(cos theta - sin theta) `
=` ((cos theta + sin theta)(cos^2 theta- cos theta sin theta + sin^2 theta))/((cos theta + sin theta)) + ((cos theta - sin theta )(cos^2 theta+ cos theta sin theta + sin^2 theta))/((cos theta - sin theta))`
=` (cos^2 theta + sin ^2 theta - cos theta sin theta ) + ( cos^2 theta + sin^2 theta + cos theta sin theta)`
=`(1- cos theta sin theta) +( 1+ cos theta sin theta)`
= 2
= RHS
Hence, LHS = RHS
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