Differentiation (Part-38) | S N De

 

$\,4.\,\,$ If $\,\,a\cos^2(C/2)+c\cos^2(A/2)=\frac{3b}{2}\,\,\,$ then show that the sides of the triangle are in A.P.

Sol. We have, $\,\,a\cos^2(C/2)+c\cos^2(A/2)=\frac{3b}{2}\\ \Rightarrow a.2\cos^2(C/2)+c.2\cos^2(A/2)=3b \\ \Rightarrow a(1+\cos C)+c(1+\cos A)=3b \\ \Rightarrow a+c+(a\cos C+c\cos A)=3b \\ \Rightarrow a+c+b=3b \\ \Rightarrow a+c=2b$

Hence, the sides of the triangle are in A.P.

$\,5.\,$ In any $\,\triangle ABC,\,$ if $\,\,\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c},\,\,$ show that, $\,\,C=60^{\circ}.$

Sol.$\,\,\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c},\,\,\text{(given)} \\ \Rightarrow \frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3 \\ \Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\ \Rightarrow \frac{b^2+bc+a^2+ac}{(a+c)(b+c)}=1 \\ \Rightarrow b^2+bc+a^2+ac=ab+ac+bc+c^2 \\ \Rightarrow a^2+b^2-c^2=ab \\ \Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac 12 \\ \Rightarrow \cos C=\frac 12=\cos 60^{\circ} \\ \Rightarrow C=60^{\circ}\,\,\,\text{(showed)}$

$\,6.\,\,$ If $\,\,2\cos A=\frac{\sin B}{\sin C}\,\,$ then show that the triangle is isosceles.

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, $\,2\cos A=\frac{\sin B}{\sin C} \\ \Rightarrow 2\sin C\cos A=\sin B \\ \Rightarrow \sin(A+C)+\sin(C-A)=\sin B \\ \Rightarrow \sin(\pi-B)+\sin(C-A)=\sin B  \\ ~~~~~~~~~~~~~~~~~[\,\,\because A+B+C=\pi] \\ \Rightarrow \sin B+\sin(C-A)=\sin B \\ \Rightarrow \sin (C-A)=0 \\ \Rightarrow C-A=0 \\ \Rightarrow 2R \sin C=2R \sin A \\ \Rightarrow c=a$

Hence, the two sides of the triangle are equal and  so the triangle is isosceles.

$\,7(i)\,$ If the cosines of two angles of a triangle are proportional to the opposite sides, show that the triangle is isosceles.

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 $\,\,\triangle ABC.$

By question, $\,\frac{\cos A}{a}=\frac{\cos B}{b} \\ \Rightarrow b\cos A=a\cos B \\ \Rightarrow 2R\sin B\cos A=2R\sin A\cos B \\ \Rightarrow 2R[\sin A\cos B-\cos A\sin B]=0 \\ \Rightarrow 2R\sin(A-B)=0 \\ \Rightarrow \sin(A-B)=0 \\ \Rightarrow A-B=0  \\ \Rightarrow 2R\sin A=2R\sin B \\ \Rightarrow a=b$ 

Hence, the two sides of the triangle are equal and  so the triangle is isosceles.

$\,(ii)\,$ In a given triangle the ratio of the cosines of two particular angles is equal to the inverse of the ratio of the corresponding opposite sides. Show that the triangle is either right angled or isosceles.

Sol. Let $\,\,a,b,c\,$ are the sides of $\,\,\triangle ABC\,\,$ which are opposite to the angles $\,A,B,C\,$ respectively.

By question, $\,\,\frac{\cos A}{1/a}=\frac{\cos B}{1/b} \\ \Rightarrow a\cos A=b\cos B \\ \Rightarrow 2R\sin A\cos A=2R\sin B\cos B \\ \Rightarrow \sin2A=\sin2B \\ \Rightarrow \sin 2A-\sin2B=0 \\ \Rightarrow 2\sin(A-B)\cos(A+B)=0  \\ \Rightarrow \sin(A-B)=0\rightarrow(1), \\ \text{or},\,\,\cos(A+B)=0 \rightarrow(2)$

From $\,(1),\,$ we get $\,\,A-B=0  \\ \Rightarrow  A=B  \\ \Rightarrow  2R\sin A=2R\sin B  \\ \Rightarrow a=b$

Again, by $(2),\,\,\,\cos(A+B)=0  \\ \Rightarrow  A+B=\pi/2  \\ \Rightarrow  \pi-C=\pi/2 \,\,[\because A+B+C=\pi] \\ \Rightarrow  C=\pi-\pi/2=\pi/2$

Hence, the triangle is either right angled or isosceles.

$\,8.\,$ In any $\,\,\triangle ABC,\,$ if $\,\,\cos A\cos B+\sin A\sin B\sin C=1\,\,$ then prove that the triangle is an isosceles right angled.

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 $\,\,\triangle ABC$

$\,\,\cos A\cos B+\sin A\sin B\sin C=1 \\ \Rightarrow 2\cos A\cos B+2\sin A\sin B\sin C=2 \\ \Rightarrow 2\cos A\cos B+2\sin A\sin B\sin C\\-(\cos^2A+\cos^2B)-(\sin^2A+\sin^2B)=0 \\ \Rightarrow (\cos^2A-2\cos A\cos B+\cos^2B)\\+(\sin^2A-2\sin A\sin B+\sin^2B)\\+2\sin A\sin B-2\sin A\sin B\sin C=0 \\ \Rightarrow (\cos A-\cos B)^2+(\sin A-\sin B)^2\\+2\sin A\sin B(1-\sin C)=0 \rightarrow(1) $

Clearly, $\,\,(\cos A-\cos B) \geq 0,\,\,(\sin A-\sin B)^2\geq 0$

Again, $\,\sin C \leq 1 \Rightarrow 1-\sin C \geq 0 \\ \text{and}\,\,\sin A >0,\,\,\sin B>0\,[\because 0<A,B <\pi]$

So, $\,\,(1)\,\,$ will be satisfied when

 $\,\, (\cos A-\cos B)^2=(\sin A-\sin B)^2=0 \\ \Rightarrow A=B \\ \Rightarrow 2R\sin A=2R \sin B \\ \Rightarrow a=b\rightarrow(2) \\ \text{and}\,\,1-\sin C=0 \\ \Rightarrow \sin C =1 \\ \Rightarrow C=\pi/2\rightarrow(3)$ 

Hence, from $\,(2),\,(3)\,$ we can conclude that the triangle is an isosceles right angled.

$\,9.\,$ In any $\,\triangle ABC,\,\,$ prove that, $\,\,\cos A+\cos B+\cos C  \leq \frac 32$

Sol. $\,\cos A+\cos B+\cos C\\=\cos A+2\cos \left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)\\=\cos A+2\cos\left(\frac{\pi}{2}-\frac A2\right)\cos\left(\frac{B-C}{2}\right)\\ ~~~~~~~~~~~~~~~~~~~~~~[\because A+B+C=\pi]\\=\cos A+2\sin (A/2)\cos\left(\frac{B-C}{2}\right)\\ \leq \cos A+2\sin(A/2)\\ ~~~~~~~~~~~~~~[\because \cos\left(\frac{B-C}{2}\right) \leq 1]\\= 1-2\sin^2(A/2)+2\sin(A/2)\\=\frac 32-2\left[\sin^2(A/2)-2.\sin(A/2).\frac 12+\frac 14\right]\\=\frac 32-2\left(\sin(A/2)-\frac 12\right)^2 \\ \leq \frac 32\quad \left[\because \left(\sin(A/2)-\frac 12\right)^2 \geq 0\right] \\ \therefore \cos A+\cos B+\cos C \leq \frac 32\,\,\text{(proved)}$