If tan a- tan b are the roots of x2-px-q-0 and a- b are the roots of x2-rx-s-0- then the value of rs is

X^2 px+q=0 tan a+ tan b=p, tan atan b=q x^2 rx+s=0 a+ b=r, a b=s Now, rs=( a+ b) a b =(1tan a+1tan b)1tan atan b = tan a+tan b(tan atan b)^2 =pq^2 nbsp; ...

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Standard XIII

Mathematics

Introduction to Angle

If tan a, tan...

Question

If $tan a, tan b$ are the roots of $x^{2} - p x + q = 0$ and $cot a, cot b$ are the roots of $x^{2} - r x + s = 0$ , then the value of $r s$ is

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Solution

$x^{2} - p x + q = 0$
$tan a + tan b = p, tan a tan b = q$
$x^{2} - r x + s = 0$
$cot a + cot b = r, cot a cot b = s$

Now,
$r s = (cot a + cot b) cot a cot b$
$= (\frac{1}{tan a} + \frac{1}{tan b}) \frac{1}{tan a tan b}$
$= \frac{tan a + tan b}{(tan a tan b)^{2}} = \frac{p}{q^{2}}$

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Introduction to Angle

Standard XIII Mathematics

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If tana,tanb are the roots of x2−px+q=0 and cota,cotb are the roots of x2−rx+s=0, then the value of rs is

x2−px+q=0 tana+tanb=p, tanatanb=q x2−rx+s=0 cota+cotb=r, cotacotb=s Now, rs=(cota+cotb)cotacotb =(1tana+1tanb)1tanatanb =tana+tanb(tanatanb)2 =pq2

If $tan a, tan b$ are the roots of $x^{2} - p x + q = 0$ and $cot a, cot b$ are the roots of $x^{2} - r x + s = 0$ , then the value of $r s$ is

$x^{2} - p x + q = 0$
$tan a + tan b = p, tan a tan b = q$
$x^{2} - r x + s = 0$
$cot a + cot b = r, cot a cot b = s$

Now,
$r s = (cot a + cot b) cot a cot b$
$= (\frac{1}{tan a} + \frac{1}{tan b}) \frac{1}{tan a tan b}$
$= \frac{tan a + tan b}{(tan a tan b)^{2}} = \frac{p}{q^{2}}$