In Delta ABC,, left( {cot frac{A}{2} + cot fr | vedclass

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(D) We know that $\cot \frac{A}{2} + \cot \frac{B}{2} = \frac{\cos(A/2)}{\sin(A/2)} + \frac{\cos(B/2)}{\sin(B/2)} = \frac{\sin(A/2+B/2)}{\sin(A/2)\sin

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$c \cot \frac{C}{2}$

Vedclass · Answer

(D) We know that $\cot \frac{A}{2} + \cot \frac{B}{2} = \frac{\cos(A/2)}{\sin(A/2)} + \frac{\cos(B/2)}{\sin(B/2)} = \frac{\sin(A/2+B/2)}{\sin(A/2)\sin(B/2)} = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)}$.Substituting this into the expression:$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( a \sin^2 \frac{B}{2} + b \sin^2 \frac{A}{2} \right)$Using the half-angle formulas $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$ and $\sin^2 \frac{B}{2} = \frac{(s-a)(s-c)}{ac}$:$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( a \frac{(s-a)(s-c)}{ac} + b \frac{(s-b)(s-c)}{bc} \right)$$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( \frac{(s-a)(s-c)}{c} + \frac{(s-b)(s-c)}{c} \right)$$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \cdot \frac{s-c}{c} (s-a+s-b) = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \cdot \frac{s-c}{c} (c)$$E = \cos \frac{C}{2} \cdot \frac{s-c}{\sin(A/2)\sin(B/2)}$Since $\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\sin(B/2) = \sqrt{\frac{(s-a)(s-c)}{ac}}$:$E = \cos \frac{C}{2} \cdot \frac{s-c}{\sqrt{\frac{(s-b)(s-c)}{bc} \cdot \frac{(s-a)(s-c)}{ac}}} = \cos \frac{C}{2} \cdot \frac{s-c}{\frac{s-c}{c} \sqrt{\frac{(s-a)(s-b)}{ab}}}$$E = c \cos \frac{C}{2} \sqrt{\frac{ab}{(s-a)(s-b)}} = c \cos \frac{C}{2} \cdot \frac{1}{\sin(C/2)} = c \cot \frac{C}{2}$.

In $\Delta ABC,\, \left( {\cot \frac{A}{2} + \cot \frac{B}{2}} \right)\,\left( {a{{\sin }^2}\frac{B}{2} + b{{\sin }^2}\frac{A}{2}} \right) =$

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