cosB-cosA=2{ a-b/c} cosC/2
Answers
Answered by
0
Answer:
Given that A,B,C are the angles of a triangle.
Therefore, A+B+C=180
Given that cosA+2cosB+cosC=2
⟹cosA+cosC=2(1−cosB)
⟹2cos(
2
A+C
)cos(
2
A−C
)=2(2sin
2
2
B
)
⟹sin(
2
B
)cos(
2
A−C
)=2sin
2
2
B
⟹cos(
2
A−C
)=2sin
2
B
Multiplying both sides by 2cos
2
B
we get 2cos
2
B
cos(
2
A−C
)=2(2cos
2
B
sin
2
B
)
⟹2cos
2
180−(A+C)
cos(
2
A−C
)=2sinB
⟹2sin(
2
A+C
)cos(
2
A−C
)=2sinB
⟹sinA+sinC=2sinB
⟹a+c=2b
Therefore, a,b,c are in A.P.
Step-by-step explanation:
hope it's helpful for you
pls mark me as brainlist with following me
Similar questions