"If $$0

Question

"If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi }$ then :"

Accepted Answer

"$x = 1 + {\cos ^2}\theta + ..........\infty$

$x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}$ .......(1)

$y = 1 + {\sin ^2}\phi + ........\infty$

$y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}$ ....... (2)

$$z = {1 \over {1 - {{\cos }^2}\theta .{{\sin }^2}\phi }} = {1 \over {1 - \left( {1 - {1 \over x}} \right)\left( {1 - {1 \over y}} \right)}} = {{xy} \over {xy - (x - 1)(y - 1)}}$$

$\Rightarrow$ $xz + yz - z = xy$

$\Rightarrow$ $xy + z = (x + y)z$"

If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi }$ then :

Solution

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If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi }$ then :

Solution

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