Question
The value of $$\tan {63^ \circ } - \cot {63^ \circ }$$ is equal to
A.
$$\frac{2}{{\sqrt 5 + 1}} \cdot \sqrt {10 - 2\sqrt 5 } $$
B.
$$\frac{2}{{\sqrt 5 + 1}} \cdot \sqrt {10 + 2\sqrt 5 } $$
C.
$$\frac{{\sqrt 5 - 1}}{4} \cdot \sqrt {10 - 2\sqrt 5 } $$
D.
None of these
Answer :
$$\frac{2}{{\sqrt 5 + 1}} \cdot \sqrt {10 - 2\sqrt 5 } $$
Solution :
Value $$ = - \frac{{\cos {{126}^ \circ }}}{{\frac{1}{2}\sin {{126}^ \circ }}} = - 2\cot {126^ \circ } = 2\tan {36^ \circ } = \frac{{4 \cdot \sin {{18}^ \circ } \cdot \cos {{18}^ \circ }}}{{\cos {{36}^ \circ }}}$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4 \cdot \frac{{\sqrt 5 - 1}}{4} \cdot \sqrt {1 - {{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)}^2}} }}{{\frac{{\sqrt 5 + 1}}{4}}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\sqrt {10 + 2\sqrt 5 } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}\left( {10 + 2\sqrt 5 } \right)} }}{{\sqrt 5 + 1}}. \cr} $$