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In △ABC,(b2−c2)cotA+(c2−a2)cotB+(a2−b2)cotC=
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REF. Image.
Given
to solve the equation
(b2−c2)cotA+(c2−a2)cotB+(a2−b2)cotC.
Let us consider properties of triangle in
triagonometry. as per that we know,
⇒asinA=bsinB=csinC=K [ where 'K' is constant]
∴ sin A = ak, sin B = bk, sin C = ck.
and also we know that,
cosA=2bcb2+c2−a2,cosB=2acc2+a2−b2,cosC=2aba2+b
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