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Question

If asin2α+bcos2α=P,bsin2β+acos2β=q,atanα=btanβ show that 1a+1b=1p+1q where ap and all of them are nonzero.

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Solution

Given asin2α+bcos2α=p....equation 1
bsin2β+acos2β=q....equation 2
a2tan2α=b2tan2β.....equation 3

Divide equation 1 by cos2α :

a2tan2α+b=psec2α

a2tan2α+b=p(1+tan2α)
tan2α=pbap

Dividing equation 2 by cos2β :
b2tan2β+a=q(1+tan2β)
tan2β=qabq

Putting values of tan2α and tan2β in equation 3:
a2(pbap)=b2(qabq)
a2[(pb)(bq)]=b2[(qa)(ap)]
a2bppqa2+a2bqa2b2=b2aqb2pq+b2apb2a2
abp(ap)+abq(ab)=pq(a2b2)
abp+abq=pq(a+b)
(p+q)ab=pq(a+b)
1p+1q=1a+1b

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