Question

$A=\tan 1,$ $B=\tan 2,$ $C=\tan 3,$ then the descending
order of $A,B,C$ is

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

Answer 1

The correct option is C $A,C,B$

We straight away see that tan2 < 0 because $\frac{\Pi }{2}<2<\Pi$

but tan1 > 0 because 0 < 1 < $\frac{\Pi }{2}$

and tan3 < 0 because $\frac{\Pi }{2}<3<\Pi$

now $tan3-tan2=\frac{sin3}{cos3}-\frac{sin2}{cos2}=\frac{sin3cos2-cos3sin2}{cos3cos2}=\frac{sin1}{cos3cos2}>0$

because $\sin 1>0and\cos 3<0and\cos 2<0$ because $\frac{\Pi }{2}<2,3<\Pi$

so $\tan 1>\tan 3>\tan 2$