Question

For 0< $\varphi$ < $\frac{\pi }{2},ifx={\sum }_{n=0}^{\infty }co{s}^{2n}\varphi ,y={\sum }_{n=0}^{\infty }si{n}^{2n}\varphi ,z={\sum }_{n=0}^{\infty }co{s}^{2n}\varphi ,then$

• A. xyz = xz + y
• B. xyz = xy + z
• C. xyz = x + y + z
• D. xyz = yz + x

Answer 1

Answer: (B), (C)

The correct options are
B

xyz = xy + z

C

xyz = x + y + z

$x=1+co{s}^2\varphi +co{s}^4\varphi +....=\frac{1}{(1-co{s}^2\varphi )}=\frac{1}{si{n}^2\varphi }y=1+si{n}^2\varphi +si{n}^4\varphi +...=\frac{1}{(1-si{n}^2\varphi )}=\frac{1}{co{s}^2\varphi }z=1+co{s}^2\varphi si{n}^2\varphi +co{s}^4\varphi si{n}^4\varphi +..=\frac{1}{(1-co{s}^2\varphi si{n}^2\varphi )}$

$Nowxyz=\frac{1}{si{n}^2\varphi co{s}^2\varphi (1-co{s}^2\varphi si{n}^2\varphi )}xy+z=\frac{1}{si{n}^2\varphi co{s}^2\varphi }+\frac{1}{1-co{s}^2\varphi si{n}^2\varphi }=\frac{1}{si{n}^2\varphi co{s}^2\varphi (1-co{s}^2\varphi si{n}^2\varphi )}=xyz$

which is given in (b)

Also x + y + z = xyz , which is given in (c).