What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))…((\sqrt{3}+\tan(29^\circ)) $$
Trigonometry – Calculating Complex Trigonometric Product Involving Square Roots and Tangents
trigonometry
Related Solutions
A proof without words (but it uses some geometry). Is that OK?
We can prove $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ (Proof below)
Putting $x=6^\circ,$ $$\tan 6^\circ\tan 66^\circ\tan 54^\circ=\tan18^\circ$$
Putting $x=18^\circ,$ $$\tan 18^\circ\tan42^\circ\tan78^\circ=\tan54^\circ$$
On multiplication, $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=1$$
$$\tan 6^\circ\tan42^\circ=\cot66^\circ\cot78^\circ=\tan(90^\circ-66^\circ)\tan(90^\circ-78^\circ)=\tan24^\circ\tan12^\circ$$
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Proof $1:$ $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan x\cdot\frac{\tan60^\circ-\tan x}{1+\tan60^\circ\tan x}\cdot\frac{\tan60^\circ+\tan x}{1-\tan60^\circ\tan x}$$
$$=\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x}\cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x}$$
$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3x$$
Proof $2:$
If $\tan3x=\tan 3y$
$$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3y$$
$$\implies \tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$
If $\tan x_1,\tan x_2,\tan x_3$ are the three roots of above cubic equation,
$\implies \tan x_1\cdot\tan x_2\cdot\tan x_3=-\tan3y$
Again, as $\tan3x=\tan 3y$
$3x=180^\circ n+3y$ where $n$ is any integer $\implies x=60^\circ n+y$
So, the in-congruent values of $x$ are $y,60^\circ +y,60^\circ\cdot2+y$
So, the corresponding roots are $\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)=\tan\{180^\circ-(60^\circ-y)\}=-\tan(60^\circ-y)$
$\implies -\tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=-\tan3y$
$\implies \tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=\tan3y$
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Best Answer
$\sqrt3+\tan(30^\circ-x)=\sqrt3+\frac{\frac{1}{\sqrt3}-\tan x}{1+ \frac{1}{\sqrt3}\tan x}$ $=\sqrt 3+\frac{1-\sqrt 3 \tan x}{\sqrt 3+\tan x}=\frac{4}{\sqrt 3+\tan x}$
$$\Longrightarrow(\sqrt 3+\tan(30^\circ-x))(\sqrt 3+\tan x)=4$$
Put $x=1,2,....,14^\circ$.
To complete, $x=15^\circ$, $(\sqrt 3+\tan(30^\circ-15^\circ))(\sqrt 3+\tan 15^\circ)=4$
$\implies (\sqrt 3+\tan 15^\circ)^2=4\implies \sqrt 3+\tan 15^\circ=2$ as $\tan 15^\circ>0$.
So, the answer should be $4^{14}\cdot 2=2^{29}$
A little generalization : Assuming $A≠n\frac{\pi}{2}$(where $n$ is any integer) so that $\cot A$ and $\tan A$ are non-zero finite,
$\cot A+ \tan(A-y)= \cot A+ \frac{\tan A-\tan y}{1+\tan A\tan y}=\frac{\cot A + \tan A}{1+\tan A\tan y}=\frac{\csc^2A}{\cot A + \tan y}$ (multiplying the numerator & the denominator by $\cot A$)
$\implies (\cot A+ \tan(A-y))(\cot A + \tan y)=\csc^2A $ , here $A=30^\circ$.