Question

In $△ABC,(b^2-c^2)\cot A+(c^2-a^2)\cot B+(a^2-b^2)\cot C$=

Answer 1

REF. Image.

Given

to solve the equation

$(b^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC.$

Let us consider properties of triangle in

triagonometry. as per that we know,

$\Rightarrow \frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=K$ [ where 'K' is constant]

$\therefore$ sin A = ak, sin B = bk, sin C = ck.

and also we know that,

$cosA=\frac{b^2+c^2-a^2}{2bc},cosB=\frac{c^2+a^2-b^2}{2ac},cosC=\frac{a^2+b^2-c^2}{2ab}$

$\because cot\theta =\frac{cos\theta }{sin\theta },$ Let us apply cot with the above propeties.

$\therefore$ we get, $cotA=\frac{b^2+c^2-a^2}{2abck},cotB=\frac{c^2+a^2-b^2}{2abck},$ and

$cotC=\frac{a^2+b^2-c^2}{2abck}$

Now, let us substtiude the above obtained values in the given equation, there by we get,

$\Rightarrow (b^2-c^2)\frac{b^2+c^2-a^2}{2abck}+(c^2-a^2)\frac{c^2+a^2-b^2}{2abck}+(a^2-b^2)\frac{a^2+b^2-c^2}{2anck}$

$\Rightarrow \frac{1}{2abck}\left[\right(b^2-c^2\left)\right(b^2+c^2-a^2)+(c^2-a^2\left)\right(c^2+a^2+b^2)+(a^2-b^2\left)\right(a^2+b^2-c^2\left)\right]$

$\therefore$ Let us simplify it further, we get

$\Rightarrow \frac{1}{2abck}[b^4+b^2{c}^2-a^2{b}^2-c^4{b}^2-c^4+a^2{c}^2+c^4+a^2{c}^2-b^2{c}^2-a^2{c}^2-a^4+a^2{b}^2+a^4+a^2{b}^2-a^2{c}^2-a^2{b}^2-b^4+b^2{c}^2]$

$\Rightarrow \frac{1}{2abck}\left(0\right)=0$

$\therefore (b^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC=0$