Eliminate θ from the relations: sec θ=1-b tan θ and a²sec²θ=5+b²tan²θ?

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Answer 1 · 2018-06-14T11:53:19

Please see the explanation below

We need

$tan^2theta+1=sec^2theta$

Therefore,

From the second relation

$a^2sec^2theta=5+b^2tan^2theta$

$a^2sec^2theta=5+b^2(sec^2theta-1)$

$a^2sec^2theta=5+b^2sec^2theta-b^2$

$sec^2theta=(5-b^2)/(a^2-b^2)$

$sectheta=sqrt((5-b^2)/(a^2-b^2))$

$tan^2theta=sec^2theta-1=(5-b^2)/(a^2-b^2)-1$

$=(5-b^2-a^2+b^2)/(a^2-b^2)$

$=(5-a^2)/(a^2-b^2)$

$tantheta=sqrt((5-a^2)/(a^2-b^2))$

Substituting in the first equation

$sectheta=1-btantheta$

$sqrt((5-b^2)/(a^2-b^2))=1-b*sqrt((5-a^2)/(a^2-b^2))$

This is the relation without $theta$