Trigonometrical Equations 3 Question 10
11. For $0<\varphi<\pi / 2$, if $x=\sum _{n=0}^{\infty} \cos ^{2 n} \varphi, y=\sum _{n=0}^{\infty} \sin ^{2 n} \varphi$, $z=\sum _{n=0}^{\infty} \cos ^{2 n} \varphi \sin ^{2 n} \varphi$, then
(1993, 2M)
(a) $x y z=x z+y$
(b) $x y z=x y+z$
(c) $x y z=x+y+z$
(d) $x y z=y z+x$
Integer Answer Type Questions
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Answer:
Correct Answer: 11. (b, c)
Solution:
- For $0<\varphi</ \pi / 2$, we have
$ x=\sum _{n=0}^{\infty} \cos ^{2 n} \varphi=1+\cos ^{2} \varphi+\cos ^{4} \varphi+\cos ^{6} \varphi+\ldots $
It is clearly a GP with common ratio of $\cos ^{2} \varphi$ which is $\leq 1$.
Hence, $x=\frac{1}{1-\cos ^{2} \varphi}=\frac{1}{\sin ^{2} \varphi} \quad \because S _{\infty}=\frac{a}{1-r},-1<r<1$
Similarly, $y=\frac{1}{\cos ^{2} \varphi}$
and
$ z=\frac{1}{1-\sin ^{2} \varphi \cos ^{2} \varphi} $
Now, $x+y=\frac{1}{\sin ^{2} \varphi}+\frac{1}{\cos ^{2} \varphi}$
$ =\frac{\cos ^{2} \varphi+\sin ^{2} \varphi}{\cos ^{2} \varphi \sin ^{2} \varphi}=\frac{1}{\cos ^{2} \varphi \sin ^{2} \varphi} $
Again, $\quad \frac{1}{z}=1-\sin ^{2} \varphi \cos ^{2} \varphi=1-\frac{1}{x y}$
$\Rightarrow \quad \frac{1}{z}=\frac{x y-1}{x y} \Rightarrow x y=x y z-z$
$\Rightarrow \quad x y+z=x y z$
Therefore, (b) is the answer from Eq. (i).
[putting the value of $x y$ ]
$\Rightarrow \quad x y z=x+y+z$
Therefore, (c) is also the answer.
