Question

If exp {$(ta{n}^{2}x+ta{n}^4+ta{n}^6+...\infty )lo{g}_{e}16$}, $0<x<\pi /4,$ satisfies the quadratic equation $x^2-3x+2=0$, then the value of $co{s}^{2}x+co{s}^{4}x$ is

• A. $\frac{4}{5}$
• B. $\frac{55}{36}$
• C. $\frac{17}{11}$
• D. $\frac{19}{31}$

Answer 1

Answer: (B)

$\frac{55}{36}$

$ta{n}^{2}x+ta{n}^{4}x+ta{n}^{6}x+...\infty =\frac{ta{n}^{2}x}{1-ta{n}^{2}x}$
Thus , ${16}^{\frac{ta{n}^{2}x}{1-ta{n}^{2}x}}$ satisfies the quadratic equation
$\Rightarrow {16}^{\frac{ta{n}^{2}x}{1-ta{n}^{2}x}}=1,2$
$\Rightarrow \frac{ta{n}^{2}x}{1-ta{n}^{2}x}=0,\frac{1}{4}$
Now, $0<x<\frac{\pi }{4}\Rightarrow 0<tanx<1$
Hence, $\frac{ta{n}^{2}x}{1-ta{n}^{2}x}=\frac{1}{4}$

$\Rightarrow 5ta{n}^{2}x=1$
$\Rightarrow co{s}^{2}x=\frac{5}{6}$
$\therefore co{s}^{2}x+co{s}^{4}x=co{s}^{2}x(1+co{s}^{2}x)=\frac{55}{36}$