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Q. In $ \Delta \,\,ABC, $ if $ s=\frac{a+b+c}{2}, $ then $ \left( b\,\,{{\cos }^{2}}\,\frac{C}{2}+c\,\,{{\cos }^{2}}\,\frac{B}{2} \right) $ is equal to

J & K CETJ & K CET 2009Trigonometric Functions

Solution:

$ b\,{{\cos }^{2}}\frac{C}{2}+c\,{{\cos }^{2}}\frac{B}{2} $
$ =b{{\left( \sqrt{\frac{s(s-c)}{ab}} \right)}^{2}}+c{{\left( \sqrt{\frac{s(s-b)}{ca}} \right)}^{2}} $
$ =b\left( \frac{s(s-c)}{ab} \right)+c\left( \frac{s(s-b)}{ca} \right) $
$ =\frac{{{s}^{2}}-sc+{{s}^{2}}-sb}{a}=\frac{2{{s}^{2}}-s(b+c)}{a} $
$ =\frac{2{{s}^{2}}-s(2s-a)}{a} $
$ =\frac{2{{s}^{2}}-2{{s}^{2}}+sa}{a}=s $