A) \[\frac{b-c}{b+c}\]
B) \[\frac{b+c}{b-c}\]
C) \[\frac{b-2c}{b+c}\]
D) None of these
Correct Answer: A
Solution :
In given right angled triangle, we have \[\cos A=c/b\] \[\Rightarrow \] \[\frac{1-{{\tan }^{2}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}=c/b\] \[\Rightarrow \] \[\frac{\left( 1+{{\tan }^{2}}\frac{A}{2} \right)-\left( 1-{{\tan }^{2}}\frac{A}{2} \right)}{\left( 1+{{\tan }^{2}}\frac{A}{2} \right)+\left( 1-{{\tan }^{2}}\frac{A}{2} \right)}=\frac{(b-c)}{(b+c)}\] (by componendo and dividendo rule) Hence, \[{{\tan }^{2}}\frac{A}{2}=\frac{b-c}{b+c}\]
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