Question

The value of $k$ for which the equation $\frac{k^2}{1-{\tan }^{2}x}=\frac{{\sin }^{2}x+k^2-2}{\cos 2x},$ where $\tan x\ne 1$ has a solution if

• A. $k>1$
• B. $k<-1$
• C. $1<k<1$
• D. $k\le -1$ or $k\ge 1$

Answer 1

The correct option is D $k\le -1$ or $k\ge 1$

$\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$
substituting in given equation,
$\frac{k^2}{1-{\tan }^{2}x}=\frac{{\sin }^{2}x+k^2-2}{\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}$
Now,
$1+{\tan }^{2}x={\sec }^{2}x$
$1-{\tan }^{2}x\ne 0\Rightarrow k^2={\sec }^{2}x({\sin }^{2}x+k^2-2)$
$k^2{\cos }^{2}x=1-{\cos }^{2}x+k^2-2$
${\cos }^{2}x(1+k^2)=k^2-1$
${\cos }^{2}x=\frac{k^2-1}{1+k^2}$
$0\le {\cos }^{2}x\le 1$
$\Rightarrow 0\le \frac{k^2-1}{1+k^2}\le 1$
$0\le k^2-1\le 1+k^2$
$\Rightarrow k^2-1\ge 0$
$\Rightarrow k\le -1$ or $k\ge 1$