MCQ
$\frac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} = $
  • A
    $\frac{1}{2}$
  • $\frac{{\sqrt 3 }}{2}$
  • C
    $\frac{{3\sqrt 3 }}{4}$
  • D
    $\sqrt 3 $

Answer

Correct option: B.
$\frac{{\sqrt 3 }}{2}$
b
(b) $\frac{{{{\cot }^2}{{15}^o} - 1}}{{{{\cot }^2}{{15}^o} + 1}} $

$= \frac{{\frac{{{{\cos }^2}{{15}^o}}}{{{{\sin }^2}{{15}^o}}} - 1}}{{\frac{{{{\cos }^2}{{15}^o}}}{{{{\sin }^2}{{15}^o}}} + 1}}$

$ = \frac{{{{\cos }^2}{{15}^o} - {{\sin }^2}{{15}^o}}}{{{{\cos }^2}{{15}^o} + {{\sin }^2}{{15}^o}}}$

$= \cos ({30^o}) = \frac{{\sqrt 3 }}{2}$.

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