Differentiation (Part-38) | S N De

 $\,(xx)\,\,\frac{\cos 2A}{a^2}-\frac{\cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}$

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, $\,\frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}\\=\frac{1-2\sin^2A}{a^2}-\frac{1-2\sin^2B}{b^2}\\=\frac{1}{a^2}-\frac{1}{b^2}+2\left[\frac{\sin^2B}{b^2}-\frac{\sin^2A}{a^2}\right]\\=\frac{1}{a^2}-\frac{1}{b^2} +2\times 0\\ \left[\because \frac{a}{\sin A}=\frac{b}{\sin B}\right]\\=\frac{1}{a^2}-\frac{1}{b^2}\,\,\,\,\text{(proved)}$

$\,(xxi)\,(a+b+c)\left(\tan(A/2)+\tan(B/2)\right)\\=2c\cot(C/2)$

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, $\,\tan(A/2)\\=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\\=\sqrt{\frac{(s-b)(s-c)(s-b)(s-c)}{s(s-a)(s-b)(s-c)}}\\=\frac{(s-b)(s-c)}{\Delta}$

Similarly, $\,\,\tan(B/2)=\frac{(s-c)(s-a)}{\Delta}$

Now, $\,\,(a+b+c)\left(\tan(A/2)+\tan(B/2)\right)\\=2s. \left[\frac{(s-b)(s-c)}{\Delta}+\frac{(s-c)(s-a)}{\Delta}\right]\\=2s.\frac{s-c}{\Delta}.(2s-b-a)\\=2s.\frac{s-c}{\Delta}.c\,\,\,[\because 2s=a+b+c]\\=2c. \frac{s(s-c)}{\Delta}\rightarrow(1)$

Again, $\,\tan(C/2)=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\ \Rightarrow \cot(C/2)\\=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}\\=\sqrt{\frac{s(s-c)(s-c)s}{(s-a)(s-b)(s-c)s}}\\=\frac{s(s-c)}{\Delta}\rightarrow(2)$

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

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

Sol. For any $\,\,\triangle ABC,\,\, (s-a)\tan(A/2)\\=(s-a).\frac{(s-b)(s-c)}{\Delta}\\=(s-b).\frac{(s-a)(s-c)}{\Delta}\\=(s-b)\tan(B/2) \rightarrow(1)\\=(s-c).\frac{(s-a)(s-b)}{\Delta}\\=(s-c)\tan(C/2)\rightarrow(2)$

From $\,(1),\,(2)\,\,$ the result follows. 

$\,(xxiii)\,\,a^2b^2c^2(\sin2A+\sin2B+\sin2C) \\ =32\Delta^3$

Sol. $\,\,a^2b^2c^2(\sin2A+\sin2B+\sin2C)\\=a^2b^2c^2[2\sin(A+B)\cos(A-B)+\sin2C]\\=a^2b^2c^2[2\sin(\pi-C)\cos(A-B) \\+2\sin C\cos C]  ~~~~~~~~~[\because A+B+C=\pi]\\=2a^2b^2c^2[\sin C\cos(A-B)\\+\sin C\cos\{\pi-(A+B)\}]\\=2a^2b^2c^2.\sin C[\cos(A-B)-\cos(A+B)]\\=a^2b^2c^2. 2\sin C. 2\sin A.\sin B\\=4a^2b^2c^2. \frac{2\Delta}{ab}.\frac{2\Delta}{bc}.\frac{2\Delta}{ac} \\ [\because \Delta=\frac 12 bc\sin A=\frac 12ca \sin B=\frac 12ab \sin C]\\=32\Delta^3\,\,\,\text{(proved)}$

$\,(xxiv)\,\,a\cos A+b\cos B+c\cos C=\frac{abc}{2R^2}$

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 $\,\,\triangle ABC$

$\,\,a\cos A+b\cos B+c\cos C\\=2R\sin A\cos A+2R\sin B\cos B\\+2R\sin C\cos C\\=R[\sin2A+\sin2B+\sin2C]\\=R.4\sin A\sin B\sin C\,\,[*]\\=4R.\frac{a}{2R}.\frac{b}{2R}.\frac{c}{2R}\\=\frac{abc}{2R^2}\,\,\text{(proved)}$

Note[*] : $\,\,\sin2A+\sin2B+\sin2C \\ =4\sin A\sin B\sin C.$

$\,(xxv)\,\,a\cos(B-C)+b\cos(C-A)+c\cos(A-B)=4.\frac{\Delta}{R}$

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 $\,\,\triangle ABC$

$\,\,a\cos(B-C)+b\cos(C-A)+c\cos(A-B)\\=2R\sin A\cos(B-C)+2R\sin B \\ \times\cos(C-A)+2R\sin C\cos(A-B)\\=R[2\sin(B+C)\cos(B-C)+2\sin(C+A)\cos(C-A)\\+2\sin(A+B)\cos(A-B)]\\=R[\sin2B+\sin2C+\sin2C \\ +\sin2A+\sin2A +\sin2B]\\=2R[\sin2A+\sin2B+\sin2C]\\=2R.4\sin A\sin B\sin C\,\,[*]\\=8R.\frac{a}{2R}.\frac{b}{2R}.\frac{c}{2R}\\=\frac{abc}{R^2}\\=\frac 4R.\left(\frac{abc}{4R}\right)\\=\frac{4\Delta}{R}\,\,[\because \Delta=\frac{abc}{4R}]$

Note[*] : $\,\,\sin2A+\sin2B+\sin2C \\ =4\sin A\sin B\sin C.$

$\,2.\,$ If $\,C=60^{\circ},\,\,$ show that, $\,2a-b=2c\cos B.$

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, $\,2a-b\\=2.2R\sin A-2R\sin B\\=2R[2\sin\{\pi-(B+C)\}-\sin B]\\=2R[2\sin(B+C)-\sin B]\\=2R[2\sin(B+60^{\circ})-\sin B]\\=2R[2\sin B\cos 60^{\circ}+2\cos B\sin60^{\circ}-\sin B]\\=2R[2.\sin B.\frac 12+2\cos B.\frac{\sqrt3}{2}-\sin B]\\=2R[\sqrt3 \cos B]\\=2\sqrt3 R\cos B\\=c.\frac{2\sqrt3 R\cos B}{c}\\=c.\frac{2\sqrt3R\cos B}{2R\sin C}\\=c.\frac{\sqrt3 \cos B}{\sin 60^{\circ}}\\=c.\frac{\sqrt3\cos B}{\sqrt3/2}\\=2c\cos B\,\,\text{(proved)}$

$\,3.\,$ If $\,b+c=2a,\,\,$ show that, $\,2\sin(A/2)=\sin(B+A/2)$

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 $\,\,\triangle ABC$

$\,\,\frac{a}{b+c}=\frac{2R\sin A}{2R(\sin B+\sin C)}\\ \Rightarrow \frac 12=\frac{2\sin(A/2)\cos(A/2)}{2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{C-B}{2}\right)}\,\,[*] \\ \Rightarrow \frac 12=\frac{\sin(A/2)\cos(A/2)}{\sin(\pi/2-A/2)\cos\left[\pi/2-(A/2+B)\right]} \,\,[**] \\ \Rightarrow \frac 12=\frac{\sin(A/2)\cos(A/2)}{\cos(A/2)\sin\left(B+A/2\right)} \\ \Rightarrow \frac 12=\frac{\sin(A/2)}{\sin\left(B+A/2\right)} \\ \Rightarrow 2\sin(A/2)=\sin(B+A/2)\,\,\text{(proved)}$

Note [*]: $\, b+c=2a$

Note[**] : $\,A+B+C=\pi$