Question

If $\tan {86}^o=m$ then $\frac{\cot 4^o-\cot {86}^o}{m+\tan 4^o}$ is

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

Answer 1

Answer: (A)

$\frac{m^2-1}{m^2+1}$

$m=\tan {86}^{\circ }$

$\therefore \frac{\cot 4^{\circ }-\cot {86}^{\circ }}{m+\tan 4^{\circ }}=\frac{\tan ({90}^{\circ }-4^{\circ })-\frac{1}{\tan {86}^{\circ }}}{m+\frac{1}{\cot 4^{\circ }}}$

$(\because \cot x=\tan ({90}^{\circ }-x)=\frac{1}{\tan x})$

$=\frac{\tan {86}^{\circ }-\frac{1}{\tan {86}^{\circ }}}{m+\frac{1}{\tan {86}^{\circ }}}$

$=\frac{m-\frac{1}{m}}{m+\frac{1}{m}}$

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

This is the required answer.