Question

If $A=\frac{\cos 9^{\circ }-\sin 9^{\circ }}{\cos 9^{\circ }+\sin 9^{\circ }},B=\frac{\cos {21}^{\circ }+\sin {21}^{\circ }}{\cos {21}^{\circ }-\sin {21}^{\circ }}$ and $C=\tan {20}^{\circ }+\tan {40}^{\circ }+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$, then the correct relation between $A,B,C$ is

• A. $A>B>C$
• B. $C>B>A$
• C. $B>C>A$
• D. $C>A>B$

Answer 1

The correct option is C $B>C>A$

$A=\frac{\cos 9^{\circ }-\sin 9^{\circ }}{\cos 9^{\circ }+\sin 9^{\circ }}\Rightarrow A=\frac{1-\tan 9^{\circ }}{1+\tan 9^{\circ }}\Rightarrow A=\frac{\tan {45}^{\circ }-\tan 9^{\circ }}{\tan {45}^{\circ }+\tan 9^{\circ }}\Rightarrow A=\tan {36}^{\circ }$

$B=\frac{\cos {21}^{\circ }+\sin {21}^{\circ }}{\cos {21}^{\circ }-\sin {21}^{\circ }}\Rightarrow B=\frac{1+\tan {21}^{\circ }}{1-\tan {21}^{\circ }}\Rightarrow B=\frac{\tan {45}^{\circ }+\tan {21}^{\circ }}{\tan {45}^{\circ }-\tan {21}^{\circ }}\Rightarrow B=\tan {66}^{\circ }$

$C=\tan {20}^{\circ }+\tan {40}^{\circ }+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$
$\because \tan \left(60°\right)=\tan (40°+20°)\Rightarrow \tan \left(60°\right)=\frac{\tan 40°+\tan 20°}{1-\tan 40°\tan 20°}\Rightarrow \tan \left(60°\right)=\tan 40°+\tan 20°+\tan \left(60°\right)\tan 40°\tan 20°\therefore C=\tan \left(60°\right)$


From the graph of $\tan x$,
$B>C>A$