Answer:
SEE STEP BY SOLUTION
Step-by-step explanation:
a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)
lhs = a / 2 (b – c )
Now use sine rule
a/ sinA = b/ sinB = c/ sinC = k
So, lhs = a / 2 (b – c )= sinA / 2(sinB -sinC)
=SinA /2* 2 cos(B+C/2) cos (B-C/2)
= sinA / 4 cos (pi/2 – A/2) cos (B-C/2)
=2 sin(A/2) cos (A/2) / 4 sin(A/2) cos (B-C/2)
= cos (A/2) / cos(B-C/2)
=cos2(A/2)/ Cos(A/2) cos(B-C/2)
= cos2(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)
=2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
= 2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
=cos2(A/2)/ (cosC-cosB)
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Answers & Comments
2tanα=3tanβ
tanα=(3/2)tanβ
L.H.S.
= tan(α-β)
= (tanα-tanβ)/(1+tanα.tanβ)
= {(3/2)tanβ-tanβ}/{1+(3/2)tanβ.tanβ}
= {(3tanβ-2tanβ)/2}/{(2+3tan²β)/2}
= tanβ/(2+3tan²β)
= (sinβ/cosβ)/{2+3(sin²β/cos²β)}
= (sinβ/cosβ)/{(2cos²β+3sin²β)/cos²β}
= {(sinβ/cosβ)×cos²β}/{2cos²β+3(1-cos²β)} [∵ sin²β+cos²β=1]
= (sinβcosβ)/(2cos²β+3-3cos²β)
= {2sinβcosβ}/{2(3-cos²β)}
= sin2β/(6-2cos²β) [∵ sin2β=2sinβcosβ]
= sin2β/(5+1-2cos²β)
= sin2β/{5-(2cos²β-1)} [∵cos2β=2cos²β-1]
= sin2β/(5-cos2β)
= R.H.S. Proved.
Answer:
SEE STEP BY SOLUTION
Step-by-step explanation:
a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)
lhs = a / 2 (b – c )
Now use sine rule
a/ sinA = b/ sinB = c/ sinC = k
So, lhs = a / 2 (b – c )= sinA / 2(sinB -sinC)
=SinA /2* 2 cos(B+C/2) cos (B-C/2)
= sinA / 4 cos (pi/2 – A/2) cos (B-C/2)
=2 sin(A/2) cos (A/2) / 4 sin(A/2) cos (B-C/2)
= cos (A/2) / cos(B-C/2)
=cos2(A/2)/ Cos(A/2) cos(B-C/2)
= cos2(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)
=2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
= 2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
=cos2(A/2)/ (cosC-cosB)