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Question
If x = cot A + cos A and y = cot A – cos A then prove that `((x-y)/(x+y))^2 + ((x-y)/2)^2=1`
Solution
LHS = `((x-y)/(x+y))^2 + ((x-y)/2)^2`
=`[((cotA+cosA)-(cotA-cosA))/((cotA+cosA)+(cotA-cosA))]^2 + [((cotA+cosA)-(cotA-cosA))/2]^2`
=`[(cotA+cosA-cotA+cosA)/(cotA+cosA+cotA-cosA)]^2 + [(cotA+cosA-cotA+cosA)/2]^2`
=`[(2cosA)/(2cotA)]^2 + [(2cosA)/2]^2`
=`[(cosA)/(((cosA)/(sinA)))]^2 + [cosA]^2`
=`[(sinA cosA)/cosA]^2 + [cosA]^2`
=`[sinA]^2 + [cosA]^2`
=`sin^2 A + cos^2 A`
=1
=RHS
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