Question

in triangle ABC prove that tan B+C/2=cotA/2​

Answer 1

Step-by-step explanation:

tan(

2

∠A+∠C

)=cot(

2

B

)

Step-by-step explanation:

$$\begin{lgathered}We \: know \: that,\\ In \: Triangle \: ABC , \\\angle A + \angle B + \angle C=180\degree\end{lgathered}$$

( Angle sum property )

$$\implies \angle A+\angle C = 180\degree - \angle B$$

Divide both sides by 2 , we get

$$\implies \frac{\angle A+ \angle C}{2}=\frac{180}{2}-\frac{B}{2}$$

$$\implies \frac{\angle A+ \angle C}{2}=90\degree -\frac{B}{2}$$

$$\implies tan\big(\frac{\angle A+ \angle C}{2}\big)=tan\big(90\degree -\frac{B}{2}\big)$$

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

/* Since ,

tan(90° - A) =cotA */

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Answer 2

Answer:

tan B+C/2=cotA/2 sum of angle triangle

tan 180-A / 2 = cot A/2

tan90-A/2 = cot A/2। (tan90- angle = cotangle)

cotA/2= cotA/2