If A, B, C are the angles of a triangle ABC then the value of tan((B + C)/2) is
a)cot B/2
b)tan A/2
c)cot C/2
d)cot A/2
Answers
Answer:
Option (d)
Step-by-step explanation:
Given :-
A, B, C are the angles of a triangle ABC.
To find :-
Find the value of Tan (B+C)/2 ?
Solution :-
Given that
In ∆ ABC , A, B and C are the three interior angles.
We know that
By Angle Sum Property
A+B+C = 180°
On dividing by 2 both sides then
=> (A+B+C)/2 = 180°/2
=> (A+B+C)/2 = 90°
=> A/2 + (B+C)/2 = 90°
=> (B+C)/2 = 90°-(A/2)
On taking Tan both sides then
=> Tan [(B+C)/2] = Tan [90°-(A/2)]
We know that
Tan (90°- θ) = Cot θ
=> Tan [(B+C)/2] = Cot (A/2)
Therefore, Tan [(B+C)/2] = Cot (A/2)
Answer:-
The value of Tan [(B+C)/2] for the given problem is Cot (A/2)
Used formulae:-
→ Tan (90°- θ) = Cot θ
→ The sum of the three angles in a triangle is 180°
Answer:
Option (d) cot A/2
Given :- A, B, C are the angles of a triangle ABC .
To Find :- The value of tan((B + C)/2) = ?
a) cot B/2
b) tan A/2
c) cot C/2
d) cot A/2
Solution :-
since A, B, C are the angles of a triangle ABC .
So,
→ A + B + C = 180° { By angle sum property. }
taking A to the RHS side,
→ B + C = 180° - A
dividing both sides by 2 ,
→ (B + C)/2 = (180° - A)/2
→ (B + C)/2 = 90° - (A/2)
taking tan both sides,
→ tan((B + C)/2) = tan(90° - (A/2))
using tan(90° - θ) = cot θ in RHS,
→ tan((B + C)/2) = cot (A/2) (d) (Ans.)
Similarly, we can also prove :-
→ tan((A + B)/2) = cot (C/2)
→ tan((A + C)/2) = cot (B/2)
Learn more :-
prove that cosA-sinA+1/cos A+sinA-1=cosecA+cotA
https://brainly.in/question/15100532
help me with this trig.
https://brainly.in/question/18213053