Question

If
${e}^{( {sin}^{2} x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 } \: satisfies \\ the \: equation \: {x}^{2} - 9x + 8 = 0 \: find \: \\ the \: value \: of \: \frac{cosx}{sinx + cosx} \: where \: 0 < x < \frac{\pi}{2}$

Please explain with proper explanation​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: {e}^{( {sin}^{2}x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 }$

We know,

$\rm \: {sin}^{2}x + {sin}^{4}x + {sin}^{6}x + - - - \infty \: forms \: an \: infinite \: gp \: with$

$\rm \: First \: term, \: a = {sin}^{2}x$

$\rm \: Common\: ratio, \: r = {sin}^{2}x$

We know,

Sum of infinite GP series with First term a and common ratio r respectively is given by

$\boxed{\tt{ S_ \infty \: = \: \frac{a}{1 - r} \: \: \: \: \:provided \: that \: |r| < 1 }}$

So, using this result, we get

$\rm \: = \: {\bigg(e \bigg) }^{\dfrac{ {sin}^{2} x}{1 - {sin}^{2} x} \times ln2}$

$\rm \: = \: {\bigg(e \bigg) }^{\dfrac{ {sin}^{2} x}{{cos}^{2} x} \times ln2}$

$\rm \: = \: {e}^{ {tan}^{2}x \: ln2}$

$\rm \: = \: {e}^{ln \: {2}^{ {tan}^{2}x } }$

$\rm \: = \: {2}^{ {tan}^{2}x }$

So,

$\rm\implies \:\boxed{\tt{ \rm \: {e}^{( {sin}^{2}x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 } = {2}^{ {tan}^{2} x}}}$

Now, Further the roots of the equation

$\rm \: {x}^{2} - 9x + 8 = 0$

$\rm \: {x}^{2} - 8x - x + 8 = 0$

$\rm \: x(x - 8) - 1(x - 8) = 0$

$\rm \: (x - 8)(x - 1) = 0$

$\rm\implies \:x = 8 \: \: \: or \: \: \: x = 1$

$\rm\implies \: {2}^{ {tan}^{2} x} = 8 \: \: \: or \: \: \: {2}^{{tan}^{2} x} = 1$

$\rm\implies \: {2}^{ {tan}^{2} x} = {2}^{3} \: \: \: or \: \: \: {2}^{{tan}^{2} x} = {2}^{0}$

$\rm\implies \:{tan}^{2} x = 3 \: \: or \: \: {tan}^{2} x = 0$

$\rm\implies \:tanx = \: \pm \: \sqrt{3} \: \: or \: \: tanx = 0$

As, it is given that

$\rm \:0 < x < \dfrac{\pi}{2}$

$\rm\implies \:tanx = \sqrt{3}$

Now, Consider

$\rm \: \dfrac{cosx}{cosx + sinx}$

can be rewritten as

$\rm \: = \: \dfrac{1}{\dfrac{cosx + sinx}{cosx} }$

$\rm \: = \: \dfrac{1}{1 + \dfrac{sinx}{cosx} }$

$\rm \: = \: \dfrac{1}{1 +tanx }$

$\rm \: = \: \dfrac{1}{1 + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{\sqrt{3} + 1} \times \dfrac{ \sqrt{3} - 1}{ \sqrt{3} - 1}$

$\rm \: = \: \dfrac{ \sqrt{3} - 1}{ {( \sqrt{3} )}^{2} - {1}^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - 1}{ 3 - 1 }$

$\rm \: = \: \dfrac{ \sqrt{3} - 1}{2 }$

Hence,

$\rm\implies \:\boxed{\tt{ \frac{cosx}{cosx + sinx} \rm \: = \: \dfrac{ \sqrt{3} - 1}{2 } \: }}$

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FORMULA USED

$\boxed{\tt{ {e}^{ln \: x} = x \: }}$

$\boxed{\tt{ y \: lnx \: = \: ln \: {x}^{y} \: }}$

Answer 2

Question:-

$\sf \large If \: {e}^{( {sin}^{2} x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 } \: satisfies \\ \sf \large the \: equation \: {x}^{2} - 9x + 8 = 0 \: find \: \\ \sf \large the \: value \: of \: \frac{cos \: x}{sin \: x + cos \: x} \: where \: 0 < x < \frac{\pi}{2} .$

Given:-

$\sf \large {e}^{( {sin}^{2} x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 }$

To Find:-

$\sf \large \frac{cos \: x}{sin \: x + cos \: x}$

Solution:-

$\sf \large Let, y \: \sf \large {e}^{ ( {sin}^{2} x + {sin}^{4}x + {sin}^{6}x + - - - \infty )ln2 }$

$\sf \large = exp \bigg \{ \frac{sin {}^{2} x}{1 - sin {}^{2}x} \bigg \}log2$

$\sf \large = exp[(tan {}^{2}x)log2 ]$

$\sf \large = exp[log2 \: tan {}^{2}x ]$

$\sf \large⇒ y = 2 \: tan {}^{2} x$

$\sf \large Now,2tan {}^{2} x \: satisfies \: the \: equation$

$\sf \large x {}^{2} - 9x + 8 = 0$

$\sf \large⇒ x = 8 \: (or) \: 1$

$\sf \large2tan {}^{2} x = 2 { }^{3} \: (or) \: 2tan {}^{2} x = 1 = 2 {}^{0}$

$\sf \large⇒ tan {}^{2} x = 3 \: (or) \: tan {}^{2} x = 0$

$\sf \large As,0 < x < \frac{\pi}{2} ,tan \: x = \sqrt{3}$

$\sf \large \therefore \: x = \frac{\pi}{3}$

$\sf \large Now, \frac{cos \: x}{cos \: x + sin \: x} = \frac{cos \frac{\pi}{3} }{cos \frac{\pi}{3} + sin \frac{\pi}{3} } = \frac{1}{1 + \sqrt{3} }$

$\sf \large = \frac{ \sqrt{3} - 1 }{2}$

Answer:-

${ \boxed{ \sf \large \blue{\sf \large \frac{cos \: x}{sin \: x + cos \: x} = \frac{ \sqrt{3} - 1 }{2} .}}}$

Hope you have satisfied. ⚘