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Question
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
Solution
LHS = `(sin theta - cos theta )/ (sin theta + cos theta) +( sin theta + cos theta )/( sin theta - cos theta )`
=` ((sin theta - cos theta )^2 + (( sin theta + cos theta )^2))/((sin theta + cos theta )( sin theta - cos theta ))`
=` (sin^2 theta + cos ^2 theta -2 sin theta cos theta + sin^2 theta + cos^2 theta + 2 sin theta cos theta)/( sin^ 2theta - cos^ 2theta)`
=` (1+1)/(sin^2 theta - ( 1-sin ^2 theta)) ( ∵ sin^2 theta + cos^2 theta =1)`
=`2/(sin^2 theta - 1 + sin^2 theta)`
=` 2/ (sin^2 theta -1)`
= RHS
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