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

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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