Question
If $$\left| {\tan A} \right| < 1,$$ and $$\left| A \right|$$ is acute then $$\frac{{\sqrt {1 + \sin 2A} + \sqrt {1 - \sin 2A} }}{{\sqrt {1 + \sin 2A} - \sqrt {1 - \sin 2A} }}$$ is equal to
A.
$$ \tan A$$
B.
$$ - \tan A$$
C.
$$ \cot A$$
D.
$$ - \cot A$$
Answer :
$$ \cot A$$
Solution :
The expression $$ = \frac{{\left| {\cos A + \sin A} \right| + \left| {\cos A - \sin A} \right|}}{{\left| {\cos A + \sin A} \right| - \left| {\cos A - \sin A} \right|}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\cos A + \sin A + \cos A - \sin A}}{{\cos A + \sin A - \left( {\cos A - \sin A} \right)}}$$ because $$ - \frac{\pi }{4} < A < \frac{\pi }{4}$$ and in this interval $$\cos A > \sin A.$$