Question

If ${\tan }^2\theta +\sec \theta =5$, find $\cos \theta$

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (C), (D)

$\frac{1}{2}$
$-\frac{1}{3}$

By replacing $sec\theta =\frac{1}{cos\theta }$ and $tan\theta =\frac{sin\theta }{cos\theta }$, we obtain the given equation as

$\frac{si{n}^2\theta }{co{s}^2\theta }+\frac{1}{cos\theta }=5$

$=\frac{si{n}^2\theta +cos\theta }{co{s}^2\theta }$

Replacing $si{n}^2\theta =1-co{s}^2\theta$, we obtain the above equation as

$1+cos\theta -co{s}^2\theta =5co{s}^2\theta$

$6co{s}^2\theta -cos\theta -1=0$, By solving the quadratic equation we obtain $cos\theta =\frac{1}{2},\frac{-1}{3}$