Question

in ∆ABC,prove that cos^2A-cos^2B/a+b + cos^2B-cos^2A/b+c +cos^2C-cos^2A/c-a =0​

Answer 1

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

We know

By sine law, In triangle ABC,

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

$\rm :\implies\:a = k \: sinA$

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

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

Consider,

$\rm :\longmapsto\:\dfrac{ {cos}^{2}A - {cos}^{2}B}{a + b}$

$\sf \: = \: \dfrac{(1 - {sin}^{2}A) - (1 - {sin}^{2}B)}{a + b}$

$\sf \: = \: \dfrac{ {sin}^{2}B - {sin}^{2}A}{k \: sinA + k \: sinB}$

$\sf \: = \: \dfrac{\cancel{(sinA + sinB)} \: (sinB - sinA)}{k \: \cancel{(sinA + sinB)}}$

$\sf \: = \: \dfrac{sinB - sinA}{k}$

$\bf\implies \:\dfrac{ {cos}^{2}A - {cos}^{2}B}{a + b} = \dfrac{sinB - sinA}{k} - - (1)$

Similarly,

$\bf\implies \:\dfrac{ {cos}^{2}B - {cos}^{2}C}{b + c} = \dfrac{sinC - sinB}{k} - - (2)$

Similarly,

$\bf\implies\:\dfrac{ {cos}^{2}C- {cos}^{2}A}{c+a} =\dfrac{sinA - sinC}{k} - - (3)$

Now,

Consider,

$\sf \:\dfrac{ {cos}^{2}A - {cos}^{2}B}{a + b} + \dfrac{ {cos}^{2}B - {cos}^{2}C}{b + c} + \dfrac{ {cos}^{2}C - {cos}^{2}A}{c + a}$

On substituting all the values evaluated above, we get

$\sf \: = \: \dfrac{sinB - sinA}{k} + \dfrac{sinC - sinB}{k} + \dfrac{sinA - sinC}{k}$

$\sf \: = \: \dfrac{sinB - sinA + sinC - sinB + sinA - sinC}{k}$

$\sf \: = \: \dfrac{0}{k}$

$\sf \: = \: 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Cosine Law :-

$\boxed{\bf\: cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2}}{2bc}}$

$\boxed{\bf\: cosB = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2}}{2ca}}$

$\boxed{\bf\: cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2}}{2ab}}$