cot⁡15∘−A+tan⁡15∘+A=

=cos⁡15∘−Asin⁡15∘−A+sin⁡15∘+Acos⁡15∘+A=cos⁡15∘−Acos⁡15∘+A+sin⁡15∘+Asin⁡15∘−Asin⁡15∘−Acos⁡15∘+A=cos⁡2A12sin⁡30∘−sin⁡2A=4cos⁡2A1−2sin⁡2A

cot⁡15∘−A+tan⁡15∘+A=

=cos⁡15∘−Asin⁡15∘−A+sin⁡15∘+Acos⁡15∘+A=cos⁡15∘−Acos⁡15∘+A+sin⁡15∘+Asin⁡15∘−Asin⁡15∘−Acos⁡15∘+A=cos⁡2A12sin⁡30∘−sin⁡2A=4cos⁡2A1−2sin⁡2A

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$\cot (15^{\circ} - A) + \tan (15^{\circ} + A) =$

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$\frac{4 \cos 2 A}{1 + 2 \cos 2 A}$

$\frac{4 \cos 2 A}{1 - 2 \sin 2 A}$

$\frac{4 \cos 2 A}{1 + 2 \sin 2 A}$

$\frac{4 \cos 2 A}{1 - 2 \cos 2 A}$

answer is B.

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Detailed Solution

$\begin{array}{l} = \frac{\cos (15^{\circ} - A)}{\sin (15^{\circ} - A} + \frac{\sin (15^{\circ} + A)}{\cos (15^{\circ} + A)} \\ = \frac{\cos (15^{\circ} - A) \cos (15^{\circ} + A) + \sin (15^{\circ} + A) \sin (15^{\circ} - A)}{\sin (15^{\circ} - A) \cos (15^{\circ} + A)} \\ = \frac{\cos 2 A}{\frac{1}{2} [\sin 30^{\circ} - \sin 2 A]} = \frac{4 \cos 2 A}{1 - 2 \sin 2 A} \end{array}$