if a b c are the angles of triangle ABC prove that tan A + c by 2 is equal to COT B by 2
Answers
Answer:
A+B+C =180 Answer:
tan\big(\frac{\angle A+ \angle C}{2}\big)=cot(\frac{B}{2})tan(
2
∠A+∠C
)=cot(
2
B
)
\begin{lgathered}We \: know \: that,\\ In \: Triangle \: ABC , \\\angle A + \angle B + \angle C=180\degree\end{lgathered}
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:
_______________________________________________________ _______________________
_________
❤
Solution:-
________
from figure;
In ∆PQR:-
<P +<Q +<R=180°. {Angle sum property}
______________
A+ B+C =180°
B+C =180°-A
___________
Dividing by 2 both side
____________
B+C/2 = 180°-A/2
B+C /2 = 90°-A/2
______________
Taken Tan both sides
________________
Tan(B+C/2)= Tan(90°-A/2)
Tan(B+C/2)=CotA/2
_______________
hence prove
hope helps ✊
