In a triangle ABC , tan A+B by 2
cot A - B by 2 is equal to
Answers
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:
Step-by-step explanation: