Question

The value of ${\prod }_{n=1}^{45}(1+\tan n^o)$ is equal to

• A. $2^{11}$
• B. $2^{22}$
• C. $2^{12}$
• D. $2^{23}$

Answer 1

Answer: (D)

$2^{23}$

First we have to know one conditional identity and that is all.

If $A+B={45}^{\circ }$

Then

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

Solution:

$=(1+\tan 1^{\circ })(1+\tan 2^{\circ })(1+\tan 3^{\circ })\dots (1+\tan {45}^{\circ })$

$=(1+\tan 1^{\circ })(1+\tan {44}^{\circ })(1+\tan 2^{\circ })(1+\tan {43}^{\circ })\dots (1+\tan {45}^{\circ })$

Clearly there are $22$ pairs of form $(1+\tan A)(1+\tan B)$

So, Total $=2^{22}\times 2=2^{23}$