Question

If $\cos x+\sec x=-2$ for a positive odd integer $n$ then ${\cos }^{n}x+{\sec }^{n}x$ is

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

Answer: (C)

$-2$

If $\cos x+\sec x=-2$

$\cos x+\frac{1}{\cos x}=-2$

${\cos }^{2}x+1=-2\cos x$
${\cos }^{2}x+1+2\cos x=0$

This is of the form $a^2+b^2+2ab=(a+b{)}^2$

$\Rightarrow (\cos x+1{)}^2=0$
$\Rightarrow (\cos x+1)=0$

$\Rightarrow \cos x=-1$

$⟹\sec x=-1$

If $n$ is odd,

${\cos }^{n}x+{\sec }^{n}x=-1+(-1)=-2$

If $n$ is even,

${\cos }^{n}x+{\sec }^{n}x=1+1=2$