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Question
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Solution
LBS = `sqrt(m/n) + sqrt(n/m)`
=`sqrt(m)/sqrt(n) + sqrt(m)/sqrt(n)`
=`(m+n)/sqrt(mn)`
=`((cos theta - sin theta ) + ( cos theta + sin theta ))/sqrt(( cos theta - sin theta ) ( cos theta + sin theta ))`
=`(2 cos theta )/ sqrt( cos ^2 theta - sin^2 theta)`
=`(2 cos theta ) / sqrt( cos ^ theta - sin^ theta)`
=` ((( 2 cos theta )/( cos theta)))/((sqrt(cos^2 theta - sin^2 theta)/(cos theta))`
=`2/(sqrt((cos^2 theta)/(cos^2 theta) - ( sin^2 theta) /( cos^2 theta))`
=`2/sqrt(1- tan^2 theta)`
= RHS
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