Question

The value of $\frac{(1+\tan {11}^{\circ })(1+\tan {34}^{\circ })}{(1+\tan {17}^{\circ })(1+\tan {28}^{\circ })}=$

• A. $1$
• B. $2$
• C. $4$
• D. None of these

Answer 1

The correct option is A $1$

If $A+B=\frac{\pi }{4}$

Then $\tan A+\tan B=1-\tan A\tan B$

$1+\tan A+\tan B+\tan A\tan B=2$

$(1+\tan A)(1+\tan B)=2$

${11}^{\circ }+{34}^{\circ }={45}^{\circ }$

$\therefore (1+\tan {11}^{\circ })(1+\tan {34}^{\circ })=2$

${17}^{\circ }+{28}^{\circ }={45}^{\circ }$

$\therefore (1+\tan {17}^{\circ })(1+\tan {28}^{\circ })=2$

$\therefore \frac{(1+\tan {11}^{\circ })(1+\tan {34}^{\circ })}{(1+\tan {17}^{\circ })(1+\tan {28}^{\circ })}=1$