Question

Please help me understand. (Evaluating the Six Trigonometric Functions)
Given that tan^2θ=3/8, what is the value of sec θ?

Answer 1

remember \(\cos^2\theta+\sin^2\theta=1\)
divide through by \(\cos^2\theta\) and you'll get \(\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}=\frac1{\cos^2\theta}\\1+\left(\frac{\sin\theta}{\cos\theta}\right)^2=\left(\frac1{\cos\theta}\right)^2\\1+\tan^2\theta=\sec^2\theta\)

Answer 2

so if we know \(\tan^2\theta=\frac38\) then \(\sec^2\theta=1+\tan^2\theta=1+\frac38=\frac{11}8\)

Answer 3

you do not have enough knowledge to determine \(\sec\theta\), however, since we do not know the sign of \(\sec\theta\) (or, equivalently, the quadrant we're dealing with)

Answer 4

this was enough, thank you! :)