Question

If $\cos x=m$ and $0<x<{90}^o$, $\tan x=$

• A. $\frac{\sqrt{1-m^2}}{m}$
• B. $\frac{m}{\sqrt{1-m^2}}$
• C. $\frac{\sqrt{1+m^2}}{m}$
• D. $\frac{m}{\sqrt{1+m^2}}$
• E. $\frac{1-m^2}{m}$

Answer 1

The correct option is A $\frac{\sqrt{1-m^2}}{m}$

Given $\cos x=m$ and $0<x<90$
Since $\cos x=m$ , we get $\sec x=\frac{1}{\cos x}=\frac{1}{m}$
$\Rightarrow \tan x=\sqrt{{\sec }^{2}x-1}$

$=\sqrt{\frac{1}{m^2}-1}$

$=\sqrt{\frac{1-m^2}{m^2}}$

$=\frac{\sqrt{1-m^2}}{m}$