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प्रश्न
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
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उत्तर
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^2 theta - cos^2 theta))`
=`(1+1)/((- cos^ 2theta )- cos^2 theta) (∵ sin^ 2theta + cos^2 theta =1)`
=`2/(1-2 cos^2 theta)`
= RHS
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