Question

if A B C are interior angles of a triangle abc then show that tan (A+B/2)=COTC/2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given,

Interior angles of triangle ABC

∠A, ∠B, ∠C

To prove,

$tan(\frac{A+B}{2} )=cot\frac{C}{2}$

Proof,

According to the angle sum property of triangles,

Sum of all interior angles of triangle = 180°

⇒∠A +∠B +∠C = $180$°

⇒∠A + ∠B = $180$° - ∠C

Now,

⇒$A+B=180-C$

dividing whole equation by 2

⇒$\frac{A+B}{2}=\frac{180-C}{2}$

⇒$\frac{A+B}{2}=90-\frac{C}{2}$

⇒$tan(\frac{A+B}{2} )=tan(90-\frac{C}{2} )$

We know that,

$tan(90-\alpha )=cot\alpha$

So, using this identity we can say that

⇒$tan(\frac{A+B}{2} )=cot(\frac{C}{2} )$

Hence proved.