Differentiation (Part-38) | S N De

 

$\,16.\,$ If $\,x,\,y,\,z\,$ are altitudes of the $\,\,\triangle ABC,\,$through the vertices $\,\,A,\,B,\,C\,\,$ respectively, then prove that, $\,\,\frac{\cos A}{x}+\frac{\cos B}{y}+\frac{\cos C}{z}=\frac 1R,\,\,$ where $\,\,R\,\,$ is the circumradius of the triangle. 

Hints : See that, $\,x=2R \sin B\sin C,\\ y=2R \sin C\sin A,\\ z=2R \sin C\sin A, \\ \sin 2A+\sin2B+\sin2C=4\sin A\sin B\sin C$

Sol.  $\,\,\frac{\cos A}{x}+\frac{\cos B}{y}+\frac{\cos C}{z}\\=\frac{\cos A}{2R \sin B\sin C}+\frac{\cos B}{2R\sin C\sin A}+\frac{\cos C}{2R\sin C\sin A}\\=\frac{1}{4R\sin A\sin B\sin C} \left(2\sin A\cos A\\ +2\sin B\cos B+2\sin C\cos C\right)\\=\frac{1}{4R\sin A\sin B\sin C}(\sin 2A+\sin2B+\sin2C)\\=\frac{1}{4\sin A\sin B\sin C} \times (4\sin A\sin B\sin C)\\=1\,\,\,\text{(proved)}$

$\,17.\,\,$ In $\,\triangle PQR,\,\,\angle R=\pi/2.\,\,$ If $\,\,\tan(P/2)\,$ and $\,\,\tan(Q/2)\,\,$ are the roots of the equation $\,\,ax^2+bx+c=0\,\,(a \neq 0),\,\,$ then show that, $\,\,a+b=c$

Sol. Since $\,\,\tan(P/2)\,$ and $\,\,\tan(Q/2)\,\,$ are the roots of the equation $\,\,ax^2+bx+c=0\,\,(a \neq 0),$

so, $ \,\, \tan(P/2)+\tan(Q/2)=-\frac ba \rightarrow(1)\\ \tan(P/2)\tan(Q/2)=\frac ca \rightarrow(2)$

Again $\,\,P+Q+R=\pi \\ \Rightarrow P/2+Q/2=\pi/2-R/2 \\ \Rightarrow \tan(P/2+Q/2)=\tan(\pi/2-\pi/4)\\~~~~[\because \angle R=\pi/2] \\ \Rightarrow \frac{\tan(P/2)+\tan(Q/2)}{1-\tan(P/2)\tan(Q/2)}=\tan(\pi/4)=1 \\ \Rightarrow \tan(P/2)+\tan(Q/2)\\ = 1-\tan(P/2)\tan(Q/2) \\ \Rightarrow -\frac ba=1-\frac ca\\ \Rightarrow -b=a-c \\ \Rightarrow c=a+b\,\,\text{(showed)}$

$\,18.\,$ In $\,\,\triangle ABC,\,$ if $\,3\left[\tan(A/2)+\tan(C/2)\right]=2\cot(B/2)\,\,$ then show that, its sides $\,\,a,\,b,\,c\,$ are in A.P.

Sol.$\,3\left[\tan(A/2)+\tan(C/2)\right]=2\cot(B/2)\,\,\text{(given)} \\ \Rightarrow 3.\left[\frac{\sin(A/2)}{\cos(A/2)}+\frac{\sin(C/2)}{\cos(C/2)}\right] =2.\frac{\cos(B/2)}{\sin(B/2)} \\ \Rightarrow 3. \left[\frac{\sin(A/2)\cos(C/2)+\cos(A/2)\sin(C/2)}{\cos(A/2)\cos(C/2)}\right] \\ =2.\frac{\cos(B/2)}{\sin(B/2)} \\ \Rightarrow 3. \frac{\sin\left(\frac{A+C}{2}\right)}{\cos(A/2)\cos(C/2)}=\frac{2\cos(B/2)}{\sin(B/2)} \\ \Rightarrow 3\sin\left(\frac{\pi-A}{2}\right)\sin(B/2) \\ =2\cos(A/2)\cos(B/2)\cos(C/2) \\ \Rightarrow 3 \times 2\cos(A/2)\sin(B/2) \\ =4\cos(A/2)\cos(B/2)\cos(C/2) \\ \Rightarrow 3\sin B=\sin A+\sin B+\sin C \\ \Rightarrow 2\sin B=\sin A+\sin C \\ \Rightarrow 2.\frac{b}{2R}=\frac{a}{2R}+\frac{c}{2R} \\ \Rightarrow 2b=a+c\rightarrow(1)$

Hence, from $\,(1)\,\,$we can conclude that  sides of the given triangle $\,\,a,\,b,\,c\,$ are in A.P.

Note : For proof of $\,\, \sin A+\sin B+\sin C \\=4\cos(A/2)\cos(B/2)\cos(C/2)\,\,$

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$\,19.\,\,$  In $\,\,\triangle ABC\,\,$ if $\,\,\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca},\,$ find $\,\angle A.$

Sol.  $\,\,\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca} \\ \Rightarrow \frac{\cos A}{a}+\left(\frac{\cos A}{a}+\frac{\cos C}{c}\right)+\left(\frac{\cos B}{b}+\frac{\cos C}{c}\right)\\=\frac{a}{bc}+\frac{b}{ca} \\ \Rightarrow \frac{\cos A}{a}+\frac{c\cos A+a\cos C}{ac}+\frac{c\cos B+b\cos C}{bc}\\=\frac{a}{bc}+\frac{b}{ca} \\ \Rightarrow \frac{\cos A}{a}+\frac{b}{ac}+\frac{a}{bc}=\frac{a}{bc}+\frac{b}{ca}\,\,[*] \\ \Rightarrow \frac{\cos A}{a}=0 \\ \Rightarrow \cos A=0 \\ \therefore A=90^{\circ}\,\,\text{(ans)}$

Note[*]: $\,\,c\cos A+a\cos C=b,\\ c\cos B+b\cos C=a.$

$\,\,20.\,\,$ In $\,\,\triangle ABC,\,C=\pi/2\,\,$ and $\,\tan(A/2),\,\tan(B/2)\,\,$ are two roots of the equation $\,\,px^2+qx+r=0\,\,(p\neq 0)\,$ then show that $\,\,p+q=r.$

Sol. Since $\,\tan(A/2),\,\tan(B/2)\,\,$ are two roots of the equation $\,\,px^2+qx+r=0\,\,(p\neq 0)\,$, we have $\,\, \tan(A/2)+\tan(B/2)=-\frac qp ,\\ \tan(A/2)\tan(B/2)=\frac rp \\ \therefore \tan(A/2+B/2)=\tan(\pi/2-C/2)\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~[\because A+B+C=\pi] \\ \Rightarrow \tan(A/2+B/2)=\tan(\pi/2-\pi/4)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\because C=\pi/2] \\ \Rightarrow \frac{\tan(A/2)+\tan(B/2)}{1-\tan(A/2)\tan(B/2)}=\tan (\pi/4)=1 \\ \Rightarrow  \tan(A/2)+\tan(B/2) \\ =1-\tan(A/2)\tan(B/2) \\ \Rightarrow -\frac qp=1-\frac rp \\ \Rightarrow p+q=r\,\,\text{(showed)}$

$\,21.\,$ In $\,\,\triangle ABC,\,\,$ if $\,\,\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13},\,\,$ prove that $\,\,\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}.$

Sol.  Let $\,\,\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k\,(\neq 0) \\ \Rightarrow b+c=11k,\rightarrow(1)\\ c+a=12k,\rightarrow(2) \\ a+b=13k\rightarrow(3) \\ \therefore 2(a+b+c)=11k+12k+13k \\ \Rightarrow a+b+c=36k/2=18k\rightarrow(4)$

Using $\,\,(1),\,(2),\,(3),\,(4)\,\,$ we get, $\,\,a=7k,\,b=6k,\,c=5k.$

Now, $\,\,\frac{\cos A}{7}=\frac 17.\frac{36+25-49}{60}=\frac{1}{35}\,[*] \\ \frac{\cos B}{19}=\frac{1}{19}.\frac{49+25-36}{70}=\frac{1}{35} \\ \frac{\cos C}{25}=\frac{1}{25}.\frac{49+36-25}{84}=\frac{1}{35}$ 

Hence follows the result.

Note [*] : $\,\,\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\,21.\,\,$ In $\,\triangle ABC,\,$ two sides $\,\,b,\,c\,$ and angle $\,B\,$ are given, side $\,a\,$ has two values $\,\,a_1,\,$ and $\,a_2\,$, show that $\,|a_1-a_2|=2\sqrt{b^2-c^2\sin^2B}$

Sol. We know, $\,\cos B=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow a^2-2ac \cos B+c^2-b^2=0 \\ \therefore a_1+a_2=2c\cos B, \\ a_1a_2=c^2-b^2 $

Now,$\,\,(a_1-a_2)^2\\=(a_1+a_2)^2-4a_1a_2\\=4c^2\cos^2B-4(c^2-b^2)\\=4b^2-4c^2(1-\cos^2B)\\=4b^2-4c^2\sin^2B\\=4(b^2-c^2\sin^2B) \\ \Rightarrow (a_1-a_2)=2\sqrt{b^2-c^2\sin^2B}\,\,\text{(showed)}$