Question

If $5 (\tan^2 x - \cos^2 x) = 2 \cos \, 2x + 9$, then the value of $\cos \, 4x$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{9}$
• C. $ - \frac{7}{9}$
• D. $ - \frac{3}{5}$

Answer 1

Answer: (C)

$5 \tan ^{2} x=9 \cos ^{2} x+7$
$5 \sec ^{2} x-5=9 \cos ^{2} x+7$
Let $\cos ^{2} x=t$
$\frac{5}{t}=9 t+12$
$9 t^{2}+12 t-5=0$
$t=\frac{1}{3}$ as $t \neq-\frac{5}{3}$
$\cos ^{2} x=\frac{1}{3}, \,\,\cos 2 x=2 \cos ^{2} x-1$
$=-\frac{1}{3}$
$\cos 4 x =2 \cos ^{2} 2 x-1$
$=\frac{2}{9}-1$
$=-\frac{7}{9}$