In triangle ABC, if a 2cos A=sin B cosec c then....
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As we know that
sinC
sinB
=
c
b
cosA=
2bc
b
2
+c
2
−a
2
Given 2cosA=sinBcosec C
⟹2cosA=
sinC
sinB
⟹2×
2bc
b
2
+c
2
−a
2
=
c
b
⟹b
2
+c
2
−a
2
=b
2
⟹c
2
=a
2
⟹c=a
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