Tangent Product series! -2

Consider the product:

$\begin{aligned} P & = (1+\tan 1^\circ) (1+\tan 2^\circ) (1+\tan 3^\circ) \cdots (1+\tan 45^\circ) \\ & = \left(1 + \frac {\sin 1^\circ}{\cos 1^\circ} \right)\left(1 + \frac {\sin 2^\circ}{\cos 2^\circ} \right)\left(1 + \frac {\sin 3^\circ}{\cos 3^\circ} \right)\cdots \left(1 + \frac {\sin 45^\circ}{\cos 45^\circ} \right) \\ & = \left(\frac {\cos 1^\circ + \sin 1^\circ}{\cos 1^\circ} \right) \left(\frac {\cos 2^\circ + \sin 2^\circ}{\cos 2^\circ} \right) \left(\frac {\cos 3^\circ + \sin 3^\circ}{\cos 3^\circ} \right) \cdots \left(\frac {\cos 45^\circ + \sin 45^\circ}{\cos 45^\circ} \right) \\ & = \frac {\sqrt 2 \cancel{\cos 44^\circ}}{\cancel{\cos 1^\circ}} \cdot \frac {\sqrt 2 \cancel{\cos 43^\circ}}{\cancel{\cos 2^\circ}} \cdot \frac {\sqrt 2 \cancel{\cos 42^\circ}}{\cancel{\cos 3^\circ}} \cdots \frac {\sqrt 2 \cancel{\cos 1^\circ}}{\cancel{\cos 44^\circ}} \cdot \frac {\sqrt 2 \cos 0^\circ}{\cos 45^\circ} \\ & = \frac {(\sqrt 2)^{45}}{\cos 45^\circ} = 2^{23} \end{aligned}$

Therefore $\log_2 P = \boxed {23}$ .

The trigonometric identity, that if A+B=45, then (1+tanA) (1+tanB)=2 can help here.

We can rearrange the interior part as follows to use the identity: (1+tan1) (1+tan44) (1+tan2) (1+tan43) ... (1+tan45)

We can thus form 22 pairs of this, as there are 44 terms which can be paired to use the identity leaving (1+tan45).

Thus, the equation will be simplified as 2 x 2 x 2 x 2 x 2 ...2 x (1+tan45) $\boxed{\text{Here 2 gets multiplied to itself 22 times, as there are 22 pairs}}$

The value of tan 45=1 and substituting this value in the equation and writing the continued 2's exponent form, we get,

${ 2 }^{ 22 }$ x (1+1)

${ 2 }^{ 22 }$ x 2

Simplifying further using the law of exponents,

${ 2 }^{ 23 }$

Now bringing the logarithm part back,

$\log _{ 2 }{ { 2 }^{ 23 } }$

• Thus, the answer is $\boxed{23}$

Let the product inside the parentheses of logarithm function be $P$ $P=(1+\tan1\degree)(1+\tan2\degree)...(1+\tan45\degree)=(1+\tan0\degree)(1+\tan1\degree)(1+\tan2\degree)...(1+\tan45\degree)$ $P=(1+\tan0\degree)(1+\tan45\degree)\times (1+\tan1\degree)(1+\tan44\degree)...$ $\blue{\because \tan0\degree=0\Rightarrow (1+\tan0\degree)=1}$ $\blue{\Rightarrow (1+\tan0\degree)(1+\tan1\degree)(1+\tan2\degree)...(1+\tan45\degree)=1\times (1+\tan1\degree)(1+\tan2\degree)...(1+\tan45\degree)=P}$ Now using a identity (proved below) ; $\boxed{(1+\tan(45\degree-x))(1+\tan x)=2}$

$\tan(45\degree-x)+1=\dfrac{1-\tan x}{1+\tan x}+1=\dfrac{1\cancel{-\tan x}+1\cancel{+\tan x}}{1+\tan x}=\dfrac{2}{1+\tan x}$ $\Rightarrow\tan(45\degree-x)+1=\dfrac{2}{1+\tan x}$ Multiplying both sides by $1+\tan x$ $(1+\tan(45\degree-x))(1+\tan x)=2$

$\Rightarrow P=2\times2\times2...\times2=2^\frac{46}{2}=2^{23}$ $\Rightarrow \log_2P=\boxed{23}$