Question

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

• A. $A,B,C$
• B. $B,A,C$
• C. $C,B,A$
• D. $B,C,A$

Answer 1

Answer: (B)

$B,A,C$

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

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

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

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

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

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

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

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

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

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

$C=-2\sqrt{2}$

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