Question

if a. b and c are interior angles of triangle ABC, then show that cot (c+a/2) = tan b/2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{a+b+c=\pi}$

$\tt{\implies\,b+c=\pi-a}$

$\tt{\implies\,\dfrac{b+c}{2}=\dfrac{\pi-a}{2}}$

$\tt{\implies\,cot\left(\dfrac{b+c}{2}\right)=cot\left(\dfrac{\pi}{2}-\dfrac{a}{2}\right)}$

$\tt{\implies\,cot\left(\dfrac{b+c}{2}\right)=tan\left(\dfrac{a}{2}\right)}$