Question

for any triangle ABC, prove that (b sq. - c sq.) cotA+ (c sq.- a sq.) cot B+ (a sq. - b sq.) cot C=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: ( {b}^{2} - {c}^{2})cotA + ( {c}^{2} - {a}^{2})cotB + ( {a}^{2} - {b}^{2})cotC$

Now, Consider first term,

$\rm \: ( {b}^{2} - {c}^{2})cotA$

By sine Law, we have

$\rm \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k$

So, we have

$\rm \: b = k \: sinB$

$\rm \: c = k \: sinC$

So, on substituting these values, we get

$\rm \: = \: ( {k}^{2} {sin}^{2}B - {k}^{2} {sin}^{2}C)cotA$

$\rm \: = \: {k}^{2} ({sin}^{2}B - {sin}^{2}C)cotA$

$\rm \: = \: {k}^{2}sin(B + C)sin(B - C)cotA$

$\rm \: = \: {k}^{2}sin(\pi - A)sin(B - C)cotA$

$\rm \: = \: {k}^{2}sinAsin(B - C) \times \frac{cosA}{sinA}$

$\rm \: = \: {k}^{2}sin(B - C)cosA$

$\rm\implies \:( {b}^{2} - {c}^{2})cotA = \: {k}^{2}sin(B - C)cosA$

Similarly,

$\rm\implies \:( {c}^{2} - {a}^{2})cotB = \: {k}^{2}sin(C - A)cosB$

Similarly,

$\rm\implies \:( {a}^{2} - {b}^{2})cotC = \: {k}^{2}sin(A - B)cosC$

On adding above three equations, we get

$\rm \: ( {b}^{2} - {c}^{2})cotA + ( {c}^{2} - {a}^{2})cotB + ( {a}^{2} - {b}^{2})cotC$

$\rm \: = {k}^{2}sin(B - C)cosA + {k}^{2}sin(C - A)cosB + {k}^{2}sin(A - B)cosC$

$\rm \: = {k}^{2}[sin(B - C)cosA + sin(C - A)cosB + sin(A - B)cosC]$

$\rm \: = {k}^{2}[(sinBcosC - sinCcosB)cosA + \\ \rm \: (sinCcosA - sinAcosC)cosB + \\ \rm \: (sinAcosB - cosAsinB)cosC]$

$\rm \: = {k}^{2}[(sinBcosCcosA - sinCcosBcosA + \\ \rm \: sinCcosAcosB - sinAcosCcosB + \\ \rm \: sinAcosBcosC - cosAsinBcosC]$

$\rm \: = \: {k}^{2} \times 0$

$\rm \: = \: 0$

Hence,

$\boxed{ \rm{ \:( {b}^{2} - {c}^{2})cotA + ( {c}^{2} - {a}^{2})cotB + ( {a}^{2} - {b}^{2})cotC = 0 }}$

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Formulae Used :-

$\boxed{ \rm{ \:sin(\pi - x) = sinx \: }}$

$\boxed{ \rm{ \:A + B + C = \pi \: }}$

$\boxed{ \rm{ \:sin(x + y)sin(x - y ) = {sin}^{2}x - {sin}^{2}y \: }}$

$\boxed{ \rm{ \:sin(x - y) = sinxcosy - sinycosx \: }}$