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STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3. trignometrical ratios,functions and identities4 Marks
MCQ
For $0 < \phi < \frac{\pi }{2},$ if $x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\phi ,} $ $y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi ,} $ $z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\phi \,{{\sin }^{2n}}\phi ,} $ then
  • A
    $xyz = xz + y$
  • B
    $xyz = xy + z$
  • C
    $xyz = x + y + z$
  • ✓
    $b$ or $c$ both
✓

Answer

Correct option: D.
$b$ or $c$ both
d
(b) $x = 1 + {\cos ^2}\phi + {\cos ^4}\phi + .... = \frac{1}{{(1 - {{\cos }^2}\phi )}} = \frac{1}{{{{\sin }^2}\phi }}$

$y = 1 + {\sin ^2}\phi + {\sin ^4}\phi + .... = \frac{1}{{(1 - {{\sin }^2}\phi )}} = \frac{1}{{{{\cos }^2}\phi }}$

$z = 1 + {\cos ^2}\phi {\sin ^2}\phi + {\cos ^4}\phi {\sin ^4}\phi + .. = \frac{1}{{(1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

Now $xyz = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

$xy + z = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi }} + \frac{1}{{1 - {{\cos }^2}\phi {{\sin }^2}\phi }}$

$ = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}} = xyz$

which is given in $(b)$

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

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