Differentiation (Part-38) | S N De

 

$\,12.\,$ Prove that, 

$\,(i)\,\,(b+c)\cos A+(c+a)\cos B \\ + (a+b)\cos C=a+b+c$

Sol. $\,\,(b+c)\cos A+(c+a)\cos B+(a+b)\cos C\\=(b\cos C+c\cos B)+(c\cos A+a\cos C)\\+(a\cos B+b\cos A)\\=a+b+c\,\,\text{(proved)}$

$\,(ii)\,\,(b-c)\sin A+(c-a)\sin B\\+(a-b)\sin C=0$

Sol. $\,\,(b-c)\sin A+(c-a)\sin B\\+(a-b)\sin C\\=(b-c)\frac{a}{2R}+(c-a)\frac{b}{2R}+(a-b)\frac{c}{2R}\\ \left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right]\\=\frac{1}{2R}(ab-ac+bc-ab+ac-bc)\\=0\,\,\text{(proved)}$

$\,\,(iii)\,\,a(b\cos C-c\cos B)=b^2-c^2$

Sol. $\,\,a(b\cos C-c\cos B)\\=ab \cos C-ac \cos B\\=ab. \frac{a^2+b^2-c^2}{2ab}-ac.\frac{a^2+c^2-b^2}{2ac}\\=\frac 12[a^2+b^2-c^2-a^2-c^2+b^2]\\=b^2-c^2\,\,\,\text{(proved)}$

$\,(iv)\,\,a^2(\cos^2B-\cos^2C)+b^2(\cos^2C\\-\cos^2A)+c^2(\cos^2A-\cos^2B)=0$

Sol. $\,\,a^2(\cos^2B-\cos^2C)+b^2(\cos^2C\\-\cos^2A)+c^2(\cos^2A-\cos^2B)\\=a^2\cos^2B-b^2\cos^2A+c^2\cos^2A \\-a^2\cos^2C+b^2\cos^2C-c^2\cos^2B\\=(a\cos B+b\cos A)(a\cos B-b\cos A)\\+(c\cos A+a\cos C)(c\cos A-a\cos C)\\+(b\cos C+c\cos B)(b\cos C-c\cos B)\\=c(a\cos B-b\cos A)+b(c\cos A-a\cos C)\\+a(b\cos C-c\cos B)\\=ac\cos B-bc \cos A+bc \cos A-ab\cos C\\+ab\cos C-ac\cos B\\=0\,\,\,\text{(proved)}$

$\,(v)\,\,2(bc \cos A+ca \cos B+ab \cos C)\\=a^2+b^2+c^2$

Sol. $\,\,2(bc \cos A+ca \cos B+ab \cos C)\\=2bc \cos A+2ca \cos B+2ab \cos C\\=2bc.\frac{b^2+c^2-a^2}{2bc}+2ca.\frac{a^2+c^2-b^2}{2ca}\\+2ab.\frac{a^2+b^2-c^2}{2ab}\\=b^2+c^2-a^2+a^2+c^2\\-b^2+a^2+b^2-c^2\\=a^2+b^2+c^2\,\,\,\text{(proved)}$

$\,(vi)\,\,(a^2-b^2-c^2)\tan A \\+(a^2-b^2+c^2)\tan B=0$

Sol. $\,\,(a^2-b^2-c^2)\tan A\\=-(b^2+c^2-a^2)\tan A \\=-\frac{abc}{R}\rightarrow(1) \\ \text{and}\,\,\,(a^2-b^2+c^2)\tan B\\=(c^2+a^2-b^2)\tan B\\=\frac{abc}{R}\rightarrow(2)$

Hence, $\,\,(a^2-b^2-c^2)\tan A+(a^2-b^2+c^2)\tan B\\=-\frac{abc}{R}+\frac{abc}{R}\\=0\,\,\text{(proved)}$

$\,\,(vii)\,\,(b^2+c^2-a^2)\tan A=(c^2+a^2\\-b^2)\tan B=(a^2+b^2-c^2)\tan C$

Sol. Clearly, $\,\,(b^2+c^2-a^2)\tan A=\frac{abc}{R}\rightarrow(1) \\(c^2+a^2-b^2)\tan B=\frac{abc}{R}\rightarrow(2) \\ (a^2+b^2-c^2)\tan C=\frac{abc}{R}\rightarrow(3)$

Hence, from $\,\,(1),\,(2),\,(3)\,\,$ it follows that

 $\,\,(b^2+c^2-a^2)\tan A=(c^2+a^2-b^2)\tan B\\=(a^2+b^2-c^2)\tan C$

$\,(viii)\,\,(b+c-a)\tan(A/2)\\=(c+a-b)\tan(B/2)\\=(a+b-c)\tan(C/2)$

Sol. $\,\,b+c-a=(a+b+c)-2a=2(s-a) \\ \therefore (b+c-a)\tan(A/2)\\=2(s-a)\times \frac{(s-b)(s-c)}{\Delta}\\=\frac{2(s-a)(s-b)(s-c)}{\Delta} \rightarrow(1)$

Similarly, we can show $\,\,(c+a-b)\tan(B/2)= \frac{2(s-a)(s-b)(s-c)}{\Delta}\rightarrow(2)\\ \,\,(a+b-c)\tan(C/2)=\frac{2(s-a)(s-b)(s-c)}{\Delta}\rightarrow(3)$

Hence, from $\,\,(1),\,(2),\,(3)\,\,$ we get, $\,\,(b+c-a)\tan(A/2)\\=(c+a-b)\tan(B/2)\\=(a+b-c)\tan(C/2)$

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