Question

The value of $A=(1+\tan 1^{\circ })(1+\tan 2^{\circ })(1+\tan 3^{\circ })\cdots (1+\tan {45}^{\circ })\text{is}$

• A. $2^{22}$
• B. $2^{23}$
• C. $2^{44}$
• D. $-2^{23}$

Answer 1

The correct option is B $2^{23}$

$\tan (45°-\theta )=\frac{1-\tan \theta }{1+\tan \theta }$
$\Rightarrow 1+\tan (45°-\theta )$$=1+\frac{1-\tan \theta }{1+\tan \theta }=\frac{2}{1+\tan \theta }$
$\Rightarrow (1+\tan (45°-\theta \left)\right)(1+\tan \theta )=2$

Put $\theta =0^{\circ },(1+\tan 45°)(1+\tan 0°)=2$
Put $\theta =1^{\circ },(1+\tan 44°)(1+\tan 1°)=2$
Put $\theta =2^{\circ },(1+\tan 43°)(1+\tan 2°)=2$
$....$
Put $\theta ={45}^{\circ },(1+\tan 0°)(1+\tan 45°)=2$
Multiplying above equations, we get
$A^2=2^{46}$
$\Rightarrow A=2^{23},-2^{23}$
But $A\ne -2^{23}$ as $\tan \theta >0$ for $1\le \theta \le {45}^{\circ }$
$\therefore A=2^{23}$