Question

(a) If x real, prove that $\frac{x^2-2x\cos \alpha +1}{x^2-2x\cos \beta +1}lies$ between $\left({\sin }^2\frac{\alpha }{2}/{\sin }^2\frac{\beta }{2}\right)$ and $\left({\cos }^2\frac{\alpha }{2}/{cos}^2\frac{\beta }{2}\right)$.
(b) Prove that the expression $\frac{tan(x+\theta )}{\tan (x-\theta )}$ does not lie between ${\tan }^2(\pi /4-\theta )$ and ${\tan }^2(\pi /4+\theta )$.
(c) Let x,y,z be real variables which satisfy the equation $xy+yz+zx=7$ and $x+y+z=6$. Find the range in which the variables lie,

Answer 1

(a) If the given expression be y, then
$x^2(y-1)-2x(ycos\beta -cos\alpha )+y-1=0$
For real values of x, $b^2-4ac\ge 0$
i.e. $4{(ycos\beta -cos\alpha )}^2-4{(y-1)}^2\ge 0$
or $\left[y\right(cos\beta -1)+(1-cos\alpha \left)\right]$
$\because$ $a^2-b^2=(a-b)(a+b)$
or $[-2y{sin}^2\frac{\beta }{2}+2{sin}^2\frac{\alpha }{2}]$
$[2y{cos}^2\frac{\beta }{2}-2{cos}^2\frac{\alpha }{2}]\ge 0$
Multiplying by $-ive$ sign and changing the sign of inequality, we get
$(y-\frac{{sin}^2\left(\alpha /2\right)}{{sin}^2\left(\beta /2\right)})(y-\frac{{cos}^2\left(\alpha /2\right)}{{cos}^2\left(\beta /2\right)})\le 0$
or $(y-a)(y-b)\le 0$ $\therefore$ $a\le y\le b$.
(b) Let $tanx=t,tan\theta =a$
$\therefore$ $tan(\frac{\pi }{4}+\theta )=\frac{1+t}{1-t},tan(\frac{\pi }{4}-\theta )=\frac{1-t}{1+t}$
where t is real.
Given expression is
$y=\frac{tanx+tan\theta }{1-tanxtan\theta }.\frac{1+tanxtan\theta }{tanx-tan\theta }$
$\frac{y}{1}=\frac{(t+a)(1+ta)}{(t-a)(1-ta)}=\frac{t(1+a^2)+a(1+t^2)}{t(1+a^2)-a(1+t^2)}$
Apply componendo and dividendo
$\frac{y+1}{y-1}=\frac{2t(1+a^2)}{2a(1+t^2)}$
$t^2\left\{a\right(y+1\left)\right\}-t(1+a^2)(y-1)+a(y+1)=0$
Since t is real $\therefore \Delta \ge 0$
${(1+a^2)}^2{(y-1)}^2-4a^2{(y+1)}^2\ge 0$
or $\left[\right(1+a^2\left)\right(y-1)+2a(y+1\left)\right]\left[\right(1+a^2\left)\right(y-1)-2a(y+1\left)\right]\ge 0$
or $[y{(1+a)}^2-{(1-a)}^2][y{(1-a)}^2-{(1+a)}^2]\ge 0$
Divide the inequality by $+ive$ quantity
${(1+a)}^2{(1-a)}^2$
or $[y-{\left(\frac{1-a}{1+a}\right)}^2][y-{\left(\frac{1+a}{1-a}\right)}^2]=+ive$.
Hence y does not lie between
${\left(\frac{1-a}{1+a}\right)}^2$ and ${\left(\frac{1+a}{1-a}\right)}^2$ where $a=tan\theta$
or ${tan}^2(\frac{\pi }{4}-\theta )$ and ${tan}^2(\frac{\pi }{4}+\theta )$
$\because$ $tan(\frac{\pi }{4}+\theta )=\frac{1+tan\theta }{1-tan\theta }=\frac{1+a}{1-a}$ etc.
(c) Eliminating z between the given relations
$xy+(x+y)z=7$
or $xy+(x+y)6-(x+y)=7$
or $xy+{(x+y)}^2+6(x+y)=7$
Arrange as a quadratic in y
$-x^2-y^2-xy+6x+6y-7=0$
or $y^2+y(x-6)+(x^2-6x+y)=0$
y being real $\Rightarrow D\ge 0$
$\therefore$ ${(x-6)}^2-4(x^2-6x+7)\ge 0$
or $3x^2-12x-8\le 0$..........(1)
or $(x-\alpha )(x-\beta )\le 0$ where $\alpha \le \beta$
$\therefore$ x lies between $\alpha$ and $\beta$ where $\alpha ,\beta$ are the roots of (1), $\alpha <\beta$
$\alpha ,\beta =\frac{6\pm 2\sqrt{15}}{3}$ $(\alpha <\beta )$
$\therefore$ $\frac{6-2\sqrt{15}}{3}<x<\frac{6+2\sqrt{15}}{2}$ $\therefore xϵ(\alpha ,\beta )$
Again x,y,z are symmetrically placed and hence y,z also lie in the same interval.