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प्रश्न
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
उत्तर
LHS = `(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))`
=`(sin theta cos theta)/(1+ sin theta - cos theta)+(cos theta sin theta)/(1+ cos theta - sin theta)`
=`sin theta cos theta [1/(1+ (sin theta - cos theta))+ 1/(1- (sin theta - cos theta))]`
=`sin theta cos theta [(1-(sin theta - cos theta)+1+(sin theta - cos theta))/({1+ (sin theta - cos theta )}{1- (sin theta-cos theta)})]`
=`sin theta cos theta[(1-sin theta + cos theta +1+sin theta - cos theta)/(1-(sin theta - cos theta)^2)]`
=`(2 sin theta cos theta)/(1-(sin ^2 theta + cos^2 theta -2 sin theta cos theta))`
=`(2 sin theta cos theta )/(2 sin theta cos theta)`
=1
= RHS
Hence, LHS = RHS
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