$\,(x)\,\,\frac{a\sin(B-C)}{b^2-c^2}=\frac{b\sin(C-A)}{c^2-a^2}=\frac{c\sin(A-B)}{a^2-b^2}$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$
Now, $\,\,\frac{a\sin(B-C)}{b^2-c^2}\\=\frac{2R\sin A\sin(B-C)}{b^2-c^2}\\=\frac{2R\sin[\pi-(B+C)]\sin(B-C)}{b^2-c^2}\\=\frac{2R\sin(B+C)\sin(B-C)}{b^2-c^2}\\=\frac{2R[\sin^2B-\sin^2C]}{b^2-c^2}\\=\frac{2R\left[\frac{b^2}{4R^2}-\frac{c^2}{4R^2}\right]}{b^2-c^2}\,\,[\text{By (1)}]\\=\frac{1}{2R} \rightarrow(2)$
Again, $\,\,\frac{b\sin(C-A)}{c^2-a^2}\\=\frac{2R\sin B\sin(C-A)}{c^2-a^2}\\=\frac{2R\sin(C+A)\sin(C-A)}{c^2-a^2}\\=\frac{2R(\sin^2C-\sin^2A)}{c^2-a^2}\\=\frac{2R\left[\frac{c^2}{4R^2}-\frac{a^2}{4R^2}\right]}{c^2-a^2}\,\,[\text{By (1)}]\\=\frac{1}{2R}\rightarrow(3)$
Similarly, $\,\,\frac{c\sin(A-B)}{a^2-b^2}=\frac{1}{2R}\rightarrow(4)$
Hence, from $\,\,(2),\,(3),\,(4)\,$, the result follows.
$\,(xi)\,\,a^2\sin2B+b^2\sin2A=4\Delta$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1),\\ \text{where R is the circumradius of}\,\,\triangle ABC$
Now, $\,\,a^2\sin2B+b^2\sin2A\\=a^2. 2\sin B\cos B+b^2. 2\sin A\cos A\\=2. a\cos B.a \sin B+2. b\cos A.b\sin A\\=2.a\cos B. b\sin A+2.b\cos A. b\sin A \\ \left[\because \frac{a}{\sin A}=\frac{b}{\sin B} \\ \Rightarrow a\sin B=b \sin A\right]\\=2b\sin A(a\cos B+b \cos A)\\=2bc \sin A\,\,\,[\because c=a\cos B+b\cos A]\\=4.\frac 12bc\sin A\\=4\Delta\,\,\,\text{(proved)}$
$\,(xii)\,\,(b+c-a)\left(\cot(B/2)+\cot(C/2)\right)\\ =2a\cot(A/2)$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1),\,\,\,$where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
Now, $\,b+c-a\\=a+b+c-2a\\=2s-2a\,\,\,[\,\because 2s=a+b+c]\\=2(s-a).$
Again, $\,\cot(B/2)=\frac{s(s-b)}{\Delta},\,\,\cot(C/2)=\frac{s(s-c)}{\Delta}.$
$\, \therefore\,(b+c-a)\left(\cot(B/2)+\cot(C/2)\right)\\=2(s-a)\left[\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta}\right]\\=2(s-a).\frac{s(2s-b-c)}{\Delta}\\=2(s-a).\frac{s(a+b+c-b-c)}{\Delta}\\=2a.\frac{s(s-a)}{\Delta}\\=2a\cot(A/2)\,\,\text{(proved)}$
$\,(xiii)\,\,a^3\sin(B-C)+b^3\sin(C-A)\\ +c^3\sin(A-B)=0$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
Now, $\,a^3\sin(B-C)\\=a^2.(2R\sin A)\sin(B-C)\\=a^2. 2R\sin[\pi-(B+C)]\sin(B-C)\\=2Ra^2\sin(B+C)\sin(B-C)\\=2R(2R\sin A)^2[\sin^2B-\sin^2C]\\=8R^3\sin^2A[\sin^2B-\sin^2C]\rightarrow(2)$
Similarly, $\,\,b^3\sin(C-A)\\=8R^3\sin^2B(\sin^2C-\sin^2A) \rightarrow(3) \\ c^3\sin(A-B)\\=8R^3\sin^2C(\sin^2A-\sin^2B) \rightarrow(4)$
Adding $\,(2),\,(3),\,(4),\,\,$ we get the required result.
$\,(xiv)\,\,b \cos B+c\cos C=a\cos(B-C)$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
Now, $\, b\cos B+c\cos C\\=2R\sin B\cos B+2R\sin C\cos C\,\,\,[\text{By (1)}]\\=R(\sin 2B+\sin2C)\\=R. 2\sin(B+C)\cos(B-C)\\=2R\sin(\pi-A)\sin(B-C)\, \\ ~~~~~~~~~~~~~~~~~~~~~[\because A+B+C=\pi]\\=2R\sin A\cos(B-C)\\=a\cos(B-C)\,\,\,\text{(proved.)}$
$\,(xv)\, bc\cos^2(A/2)+ca\cos^2(B/2)\\ + ab\cos^2(C/2)=s^2$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
Now, $\,\,bc\cos^2(A/2)+ca\cos^2(B/2)+ab\cos^2(C/2)\\=bc.\frac{s(s-a)}{bc}+ca.\frac{s(s-b)}{ca}+ab.\frac{s(s-c)}{ab}\\=s[s-a+s-b+s-c]\\=s[3s-(a+b+c)]\\=s[3s-2s]\,\,[\because 2s=a+b+c]\\=s.s\\=s^2\,\,\text{(proved)}$
$\,(xvi)\,\,\frac{\cos A}{a}+\frac{a}{bc}=\frac{\cos B}{b}+\frac{b}{ca}=\frac{\cos C}{c}+\frac{c}{ab}$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
$\,\,\frac{a}{bc}-\frac{b}{ca}\\=\frac{b\cos C+c\cos B}{bc}-\frac{a\cos C+c\cos A}{ca}\\=\frac{\cos C}{c}+\frac{\cos B}{b}-\frac{\cos C}{c}-\frac{\cos A}{a}\\ \Rightarrow \frac{\cos A}{a}+\frac{a}{bc}=\frac{\cos B}{b}+\frac{b}{ca}\rightarrow(1)$
Again, $\,\,\frac{b}{ca}-\frac{c}{ab}\\=\frac{a\cos C+c\cos A}{ca}-\frac{a\cos B+b\cos A}{ab}\\=\frac{\cos C}{c}+\frac{\cos A}{a}-\frac{\cos B}{b}-\frac{\cos A}{a}\\ \Rightarrow \frac{\cos B}{b}+\frac{b}{ac}=\frac{\cos C}{c}+\frac{c}{ab} \rightarrow(2)$
From $\,(1)\,$ and $\,(2)\,$, we get $\,\frac{\cos A}{a}+\frac{a}{bc}=\frac{\cos B}{b}+\frac{b}{ca}=\frac{\cos C}{c}+\frac{c}{ab}$
$\,(xvii)\,a^2\cot A+b^2\cot B+c^2\cot C=4\Delta$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
Now, $\,a^2\cot A+b^2\cot B+c^2\cot C\\=(2R \sin A)^2.\frac{\cos A}{\sin A}+(2R\sin B)^2.\frac{\cos B}{\sin B}\\+(2R\sin C)^2.\frac{\cos C}{\sin C}\\=2R^2[2\sin A\cos A+2\sin B\cos B\\+2\sin C\cos C]\\=2R^2[\sin 2A+\sin 2B+\sin2C]\\=2R^2[2\sin(A+B)\cos(A-B)\\+2\sin C\cos C]\\=2R^2[2\sin(\pi-C)\cos(A-B)\\+2\sin C\cos \{\pi-(A+B)\}]\\=2R^2[2\sin C\cos (A-B)\\-2\sin C\cos(A+B)]\\=2R^2.2\sin C[\cos(A-B)\\-\cos(A+B)]\\=4R^2.\sin C. 2\sin A\sin B\\=4R^2.\frac{c}{2R}.2\frac{a}{2R}.\frac{b}{2R}\\=\frac{abc}{R}\\=4\Delta\,\,[\because R=\frac{abc}{4\Delta}]$
$\,(xviii)\,\,a^2+b^2+c^2 \\=4\Delta(\cot A+\cot B+\cot C)$
Sol. We know, for any $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC.$
We know, $\,\tan A=\frac{abc}{R}.\frac{1}{b^2+c^2-a^2} \\ \Rightarrow \cot A=\frac{R}{abc}.(b^2+c^2-a^2)\rightarrow(1)$
Similarly, $\,\cot B=\frac{R}{abc}.(a^2+c^2-b^2)\rightarrow(2)\\ \cot C=\frac{R}{abc}.(a^2+b^2-c^2)\rightarrow(3) \\ \therefore 4\Delta(\cot A+\cot B+\cot C)\\=\frac{abc}{R}\left[\frac{R}{abc}(b^2+c^2-a^2\\+a^2+c^2-b^2+a^2+b^2-c^2)\right] \\ [\text{By (1),(2), (3)}]\\=a^2+b^2+c^2\,\,\text{(proved)}$
$\,(xix)\,\,\frac{\cos B\cos C}{bc}+\frac{\cos C\cos A}{ca}\\+\frac{\cos A\cos B}{ab}=\frac{\sin^2A}{a^2}$
Sol. $,\,\frac{\cos B\cos C}{bc}+\frac{\cos C\cos A}{ca}+\frac{\cos A\cos B}{ab}\\=\frac{\cos C}{c}\left[\frac{\cos B}{b}+\frac{\cos A}{a}\right]+\frac{\cos A\cos B}{ab}\\=\frac{\cos C}{c}.\frac{a\cos B+b\cos A}{ab}+\frac{\cos A\cos B}{ab}\\=\frac{\cos C}{c}.\frac{c}{ab}+\frac{\cos A\cos B}{ab}\\=\frac{1}{ab}[\cos C+\cos A\cos B]\\=\frac{1}{ab}[\cos\{\pi-(A+B)\}\\+\frac 12\{\cos(A+B)+\cos(A-B)\}]\\=\frac{1}{ab}[-\cos(A+B)\\+\frac 12 \cos(A+B)+\frac 12\cos(A-B)]\\=\frac{1}{2ab}[\cos(A-B)-\cos(A+B)]\\=\frac{1}{2ab}\times 2\sin A\sin B\\=\frac{\sin A}{a} \times \frac{\sin B}{b}\\=\frac{\sin^2 A}{a^2}\,\,[\because \frac{a}{\sin A}=\frac{b}{\sin B}]\,\,\text{(proved)}$
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