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