Help with a trig proof please?

Question

$(a tan u + b)^2+(btanu-a)^2=(a^2+b^2)sec^2u$

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Answer 1 · 2018-03-22T09:32:39

Here you go....

$(atanu+b)^2+(btanu-a)^2=(a^2+b^2)sec^2u$

$=a^2tan^2u+2abtanu+b^2+b^2tan^2u+a^2-2abtanu$

$=a^2(tan^2u+1)+b^2(tan^2u+1)$

$=a^2*sec^2u+b^2*sec^2u$

$=sec^2u(a^2+b^2)$

$=(a^2+b^2)sec^2u$

hope that helped!!