Question

For $0 < \phi < \dfrac{\pi }{2}$,if $x = \sum\nolimits_{n = 0}^\infty {{{\cos }^{2n}}} \phi$, $y = \sum\nolimits_{n = 0}^\infty {{{\sin }^{2n}}} \phi$ and $z = \sum\nolimits_{n = 0}^\infty {{{\cos }^{2n}}} \phi {\sin ^{2n}}\phi$ then $xyz$=
A) $xy + z$
B) $xz + y$
C) $x + y + z$
D) $yz + x$

Answer 1

In the given question, we are given that let $0 < \phi < \dfrac{\pi }{2}$ Also we are given the values of three variables $x$, $y$ and $z$in the terms of trigonometric terms or trigonometric ratios $\sin \phi$and $\cos \phi$ with their powers up to $n$ and therefore we have to find out the relationship between $x$, $y$ and $z$.

Complete step-by-step answer:
In the given question, we are given the range of angle $\phi$ which is $0 < \phi < \dfrac{\pi }{2}$ Also we are given the values of three variables $x$, $y$ and $z$in terms of n and $2n$ powers of trigonometric ratios $\cos \phi$ and $\sin \phi$ . Therefore using various trigonometric identities we will find out the relationship between these three variables $x$, $y$ and $z$. Firstly we are going to solve the values of $x,y,z$.
On solving $x$, we get
Given $x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}} \phi$
On substituting the values of $n = 0$to $n = \infty$and then summing all the terms because of (MISSING A SIGN)
Therefore we get,
$x = {\cos ^{2(0)}}\phi + {\cos ^{2(1)}}\phi + {\cos ^{2(2)}}\phi + .........\infty$
$x = {\cos ^0}\phi + {\cos ^2}\phi + {\cos ^4}\phi + .........\infty$
Here, using T-ratios formulas, we know, the values of ${\cos ^0}\phi = 1$
Therefore on substituting, we get
$x = 1 + {\cos ^2}\phi + {\cos ^4}\phi + .............\infty$
This is geometric progression with first term $\left( a \right){\text{ }} = {\text{ }}1$and common ratio $\left( r \right) = \dfrac{{{{\cos }^2}\phi }}{1} = {\cos ^2}\phi$. Therefore sum of geometric progression $u\dfrac{a}{{1 - r}}$
$x = \dfrac{1}{{1 - {{\cos }^2}\phi }}$
because $a = 1$& $r = {\cos ^2}\phi$
Also, ${\cos ^2}\phi + {\sin ^2}\phi = 1$ which implies ${\cos ^2}\phi = 1 - {\sin ^2}\phi$
Therefore = $\dfrac{1}{{1 - (1 - {{\sin }^2}\phi )}} = \dfrac{1}{{1 - 1 + {{\sin }^2}\phi }} = \dfrac{1}{{{{\sin }^2}\phi }}$
$x = \dfrac{1}{{{{\sin }^2}\phi }}$
On solving $y$, we get
$y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi }$
Putting values of $n = 1,2,3,.......\infty$ and summing, we get
$y = {\sin ^{2(0)}}\phi + {\sin ^{2(1)}}\phi + {\sin ^{2(2)}}\phi + ........\infty$
$y = 1 + {\sin ^2}\phi + {\sin ^4}\phi + ........\infty$Y = ...
where ${\sin ^0}\phi = 1$ and this becomes a geometric progression with $a = 1$and $r = {\sin ^2}\phi$.
so, $y = \dfrac{1}{{1 - {{\sin }^2}\phi }} = \dfrac{1}{{{{\cos }^2}\phi }}$
because ${\sin ^2}\phi + {\cos ^2}\phi = 1$ which implies $1 - {\sin ^2}\phi = {\cos ^2}\phi$.
$y = \dfrac{1}{{{{\cos }^2}\phi }}$
Also, on solving $z$, we get
$z = \sum\limits_{n = 0}^\infty {{{\cos }^2}\phi } {\sin ^2}\phi$
On putting $n = 0$to $\infty$and summing, we get
$z = {\cos ^{2(0)}}\phi {\sin ^{2(0)}}\phi + {\cos ^{2(1)}}\phi {\sin ^{2(1)}}\phi + {\cos ^{2(2)}}\phi {\sin ^{2(2)}}\phi + ........\infty$
$z = 1 + {\cos ^{2(1)}}\phi {\sin ^{2(1)}}\phi + {\cos ^{2(2)}}\phi {\sin ^{2(2)}}\phi + ........\infty$
Where ${\cos ^0}\phi {\sin ^0}\phi = 1$ and this becomes
Progression with $a = 1$ & $r = {\cos ^2}\phi {\sin ^2}\phi$
$z = \dfrac{1}{{1 - {{\cos }^2}\phi {{\sin }^2}\phi }}$
We got three values of variables $x$, $y$ and $z$
Here, we have to find $xyz$and on substituting the values of $x$, $y$ and $z$ in $xyz$ we get
$xyz = \dfrac{1}{{{{\sin }^2}\phi }}.\dfrac{1}{{{{\cos }^2}\phi }}.\dfrac{1}{{1 - {{\cos }^2}\phi {{\sin }^2}\phi }}$
$xyz = \dfrac{1}{{({{\sin }^2}\phi {{\cos }^2}\phi )(1 - {{\sin }^2}\phi {{\cos }^2}\phi )}}$
By multiplying the terms we get the above equation.
Here in the numerator, we will add and subtract the term ${\sin ^2}\phi {\cos ^2}\phi$ so we get
$xyz = \dfrac{{1 - {{\sin }^2}\phi {{\cos }^2}\phi + {{\sin }^2}\phi {{\cos }^2}\phi }}{{({{\sin }^2}\phi {{\cos }^2}\phi )(1 - {{\sin }^2}\phi {{\cos }^2}\phi )}}$
On splitting the terms in numerator, we get
$xyz = \dfrac{{1 - {{\sin }^2}\phi {{\cos }^2}\phi }}{{({{\sin }^2}\phi {{\cos }^2}\phi )(1 - {{\sin }^2}\phi {{\cos }^2}\phi )}} + \dfrac{{{{\sin }^2}\phi {{\cos }^2}\phi }}{{({{\sin }^2}\phi {{\cos }^2}\phi )(1 - {{\sin }^2}\phi {{\cos }^2}\phi )}}$
On canceling the terms we get
$xyz = \dfrac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi }} + \dfrac{1}{{1 - {{\sin }^2}\phi {{\cos }^2}\phi }}$
$xyz = \dfrac{1}{{{{\sin }^2}\phi }}\dfrac{1}{{{{\cos }^2}\phi }} + \dfrac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi }}$
On substituting the values of $x,y,z$ in this we get
$xyz = xy + z$

So, the correct answer is “Option A”.

Note: In the above question, we had solved $x$, $y$ and $z$ firstly by substituting values of $n$ from $0{\text{ }}to{\text{ }}\infty$ and then summing all the terms and then that becomes geometric progression with the common term and common ratio geometric progression is the series which is the summation of terms with one common multiple which on further divided get common ratio.