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Question

If $$A = \sin 15^{\circ} + \cos 15^{\circ}$$, $$B = \tan 15^{\circ} + \cot 15^{\circ}$$, $$C = \tan {22} \dfrac {1}{2}^{\circ} - \cot {22} \dfrac {1}{2}^{\circ}$$ then the descending order is 


A
A,B,C
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B
B,A,C
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C
C,B,A
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D
B,C,A
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Solution

The correct option is C $$B,A,C$$
$$A = \sin {15^{\circ}} + \cos {15^{\circ}}$$

$$A = \sqrt {2} \left (\dfrac {1}{\sqrt {2}}\sin {15^{\circ}}+\dfrac {1}{\sqrt {2}}\cos {15^{\circ}}\right )$$

$$\boxed {A = \sqrt {2}\sin {60^{\circ}}=\sqrt {\dfrac {3}{2}}}$$

$$B = \tan {15^{\circ}} + \cot {15^{\circ}}$$

$$=\dfrac {\sin {15^{\circ}}}{\cos {15^{\circ}}}+\dfrac {\cos {15^{\circ}}}{\sin {15^{\circ}}}$$

$$=\dfrac {2}{2\sin {15^{\circ}}\cos {15^{\circ}}}$$

$$\boxed {B=\dfrac {2}{\sin {30^{\circ}}}=4}$$

$$C=\tan {22}\dfrac {1}{2}^{\circ}-\cot {22}\dfrac {1}{2}^{\circ}$$

$$=\dfrac {\sin {22}\dfrac {1}{2}^{\circ}}{\cos {22}\dfrac {1}{2}^{\circ}}-\dfrac {\cos {22}\dfrac {1}{2}^{\circ}}{\sin {22}\dfrac {1}{2}^{\circ}}$$

$$=-\dfrac {2}{2\sin {22}\dfrac {1}{2}^{\circ}\cos {22}\dfrac {1}{2}^{\circ}}$$

$$\boxed {C=-2\sqrt2}$$

Hence, the descending order is $$B, A, C$$.

Mathematics

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