given that $\tan^{2}\theta = \frac{3}{8}$, what is the value of $sec\theta$?\n$pmsqrt{\frac{8}{3}}$\n$pmsqrt{…

given that $\tan^{2}\theta = \frac{3}{8}$, what is the value of $sec\theta$?\n$pmsqrt{\frac{8}{3}}$\n$pmsqrt{\frac{11}{8}}$\n$\frac{11}{8}$\n$\frac{8}{3}$

given that $\tan^{2}\theta = \frac{3}{8}$, what is the value of $sec\theta$?\n$pmsqrt{\frac{8}{3}}$\n$pmsqrt{\frac{11}{8}}$\n$\frac{11}{8}$\n$\frac{8}{3}$

Answer

Explanation:

Step1: Recall the trigonometric identity

We know the identity $1 + \tan^{2}\theta=\sec^{2}\theta$.

Step2: Substitute the given value

Given $\tan^{2}\theta=\frac{3}{8}$, then $\sec^{2}\theta=1+\frac{3}{8}=\frac{8 + 3}{8}=\frac{11}{8}$.

Step3: Solve for $\sec\theta$

Taking the square - root of both sides, $\sec\theta=\pm\sqrt{\frac{11}{8}}$.

Answer:

$\pm\sqrt{\frac{11}{8}}$