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sec(theta) = 9 find the value of tan^2(theta)
@jdoe0001
tan^2(theta) + sec^2(theta) = 1 tan^2(theta) = 1 - sec^2(theta) I get lost after that
@SithsAndGiggles
\(\bf 1+tan^2(\theta)=sec^2(\theta)\implies tan^2(\theta)=sec^2(\theta)-1 \\ \quad \\ \quad \\ sec(\theta)={\color{brown}{ 9}} \\ \quad \\ tan^2(\theta)=sec^2(\theta)-1\implies [tan(\theta)]^2=[sec(\theta)]^2-1 \\ \quad \\\ [tan(\theta)]^2=[{\color{brown}{ 9}}]^2-1\)
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thanks again bro! i really appreciate it. I think i understand this section now!
damn my computer is screwed up. It didnt post my latest reply where i said thanks! youre a lifesaver. i finally understand this section now
yw
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