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

Question

1070 views around the world

You can reuse this answer
Creative Commons License

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#