# IN ANY TRIANGLE PROVE THAT (a+b+c) (tanA/2+tanB/2)=2c cotc/2?

$2s=a+b+c We know, \tan (A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} Similarly, \tan (B/2) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} Now, (a+b+c)(\tan (A/2)+\tan (B/2)=(a+b+c)(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}) =2s\sqrt{\frac{s-c}{s}}(\sqrt{\frac{s-b}{s-a}}+\sqrt{\frac{s-a}{s-b}}) =2s\sqrt{\frac{s-c}{s}}(\frac{s-b+s-a}{\sqrt{(s-b)(s-a}}) =2c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} =2c\tan (C/2)$2s=a+b+c" id="MathJax-Element-741-Frame" role="presentation" style="box-sizing: border-box; margin: 0px; padding: 0px; border: 0px; outline: 0px; font-size: 14px; vertical-align: middle; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial; font-family: proxima_nova_rgregular, "helvetica neue", helvetica, arial, sans-serif; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;" tabindex="0">$2s=a+b+c We know, \tan (A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} Similarly, \tan (B/2) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} Now, Q=(a+b+c)(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}) =2s\sqrt{\frac{s-c}{s}}(\sqrt{\frac{s-b}{s-a}}+\sqrt{\frac{s-a}{s-b}}) =2s\sqrt{\frac{s-c}{s}}(\frac{s-b+s-a}{\sqrt{(s-b)(s-a}}) =2c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} =2c\tan (C/2)$$2s=a+b+c We know, \tan (A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} Similarly, \tan (B/2) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} Now, Q=(a+b+c)(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}) =2s\sqrt{\frac{s-c}{s}}(\sqrt{\frac{s-b}{s-a}}+\sqrt{\frac{s-a}{s-b}}) =2s\sqrt{\frac{s-c}{s}}(\frac{s-b+s-a}{\sqrt{(s-b)(s-a}}) =2c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} =2c\tan (C/2)$