Question

prove that tanB+c/2=cotA/2​

Answer 1

Explanation:

Answer:

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

Step-by-step explanation:

We \: know \: that,\\</p><p>In \: Triangle \: ABC ,</p><p>\\\angle A + \angle B + \angle C=180\degree

( 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 */