A) \[{{\left( \frac{1+\tan x}{1-\tan x} \right)}^{2}}\]
B) \[{{\left( \frac{1-\tan x}{1+\tan x} \right)}^{2}}\]
C) \[\frac{2{{\sin }^{2}}\frac{x}{2}}{1+{{\cos }^{2}}\frac{x}{2}}\]
D) None of these
Correct Answer: A
Solution :
\[\tan \,\,(45{}^\circ +x)=\frac{\tan 45{}^\circ +\tan x}{1-\tan 45{}^\circ tanx}=\frac{1+\tan x}{1-\tan x}\] \[\tan \,\,(45{}^\circ -x)=\frac{\tan 45{}^\circ -\tan x}{1+\tan 45{}^\circ tanx}=\frac{1-\tan x}{1+\tan x}\] \[\therefore \]\[\frac{\tan \,\,(45{}^\circ +x)}{\tan \,\,(45{}^\circ -x)}=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}={{\left[ \frac{1+\tan x}{1-\tan x} \right]}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
