Question

In a triangle ABC , tan A+B by 2
cot A - B by 2 is equal to​

Answer 1

Answer:

Answer:

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

2

∠A+∠C

)=cot(

2

B

)

Step-by-step explanation:

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

Weknowthat,

InTriangleABC,

∠A+∠B+∠C=180°

( Angle sum property )

\implies \angle A+\angle C = 180\degree - \angle B⟹∠A+∠C=180°−∠B

Divide both sides by 2 , we get

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

2

∠A+∠C

=

2

180

−

2

B

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

2

∠A+∠C

=90°−

2

B

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

2

∠A+∠C

)=tan(90°−

2

B

)

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

2

∠A+∠C

)=cot(

2

B

)

/* Since ,

tan(90° - A) =cotA */

Answer 2

Answer:

$\huge\underlined{\sf{tan (\frac{∠A+∠C}{2})=cot{B}{2}$

Step-by-step explanation: