Must find tan 5

$\displaystyle \prod_{i=1}^{5} [\sqrt{3}+ \tan (5i)^{\circ}] = \displaystyle \prod_{i=1}^{5} \frac{\sqrt{3} \cos (5i)^{\circ} + \sin (5i)^{\circ}}{\cos (5i)^{\circ}}$

$= \displaystyle \prod_{i=1}^{5} [2 \frac{\frac{\sqrt{3}}{2} \cos (5i)^{\circ} + \frac{1}{2} \sin (5i)^{\circ}}{\cos (5i)^{\circ}}] = \displaystyle \prod_{i=1}^{5} [2 \frac{\sin (60 + 5i)^{\circ}}{\cos (5i)^{\circ}}]$

$= 2^5 * \frac{\sin 65^{\circ}}{\cos 5^{\circ}} \frac{\sin 70^{\circ}}{\cos 10^{\circ}} \frac{\sin 75^{\circ}}{\cos 15^{\circ}} \frac{\sin 80^{\circ}}{\cos 20^{\circ}} \frac{\sin 85^{\circ}}{\cos 25^{\circ}} = 32$ .

All other terms cancel as $\sin x^{\circ} = \cos (90 - x)^{\circ}$ .

Note that for any angle $\theta$ , $\frac{\tan\theta+\tan(30^\circ-\theta)}{1-\tan\theta\tan(30^\circ-\theta)}=\tan(\theta+30^\circ-\theta)=\frac{\sqrt3}{3}$ $\sqrt3[\tan\theta+\tan(30^\circ-\theta)]=1-\tan\theta\tan(30^\circ-\theta).$ Therefore, $[\sqrt3+\tan\theta][\sqrt3+\tan(30^\circ-\theta)]$ $=3+\sqrt3[\tan\theta+\tan(30^\circ-\theta)]+\tan\theta\tan(30^\circ-\theta)$ $=3+1$ $=4.$ We can pair up the $5^\circ$ term with the $25^\circ$ term and the $10^\circ$ term with the $20^\circ$ term, leaving the $15^\circ$ term. But from the tangent half-angle formula or tangent angle difference formula, we have $\tan15^\circ=2-\sqrt3$ . Thus, the product is equal to $4^2\cdot2=\boxed{32}$ .

If $A + B = 30^\circ$ , then we have:

$\tan(A+B) = \frac 1 {\sqrt 3} \implies \tan A + \tan B = \frac 1 {\sqrt 3}(1 - \tan A \tan B).$

This gives us:

$(\sqrt 3 + \tan A)(\sqrt 3 + \tan B) = 3 + \sqrt 3(\tan A + \tan B) + \tan A \tan B = 4.$

Now denoting the product in the question by $A$ , we have:

$\begin{aligned}A^2 &= \prod_{i=1}^5 (\sqrt 3 + \tan(5i)^\circ) \prod_{j=1}^5(\sqrt 3 + \tan(5j)^\circ) \\ &= \prod_{i=1}^5 (\sqrt 3 + \tan(5i)^\circ) (\sqrt 3 + \tan(5(6-i))^\circ) \\ &= 4^5. \end{aligned}$

since $5i + 5(6-i) = 30$ . This gives $A = 2^5 = 32$ .

$(\sqrt{3} + tan\theta)(\sqrt{3} + tan(30^{\circ} - \theta))$

= $(\sqrt{3} +tan\theta)( \sqrt{3} + \frac{\frac{1}{\sqrt{3}} - tan\theta}{1 + \frac{1}{\sqrt{3}}tan\theta})$

= $(\sqrt{3} + tan\theta)(\frac{4}{\sqrt{3} + tan\theta }) = 4$

Now, $\displaystyle \prod_{i = 1}^{5}[\sqrt{3} + tan(5i^{\circ}) ]$

= $4 \times 4 \times (\sqrt{3} + tan(15^{\circ}))$ (try to replace $\theta = 5^{\circ}$ and $10^{\circ}$ )

= $4 \times \ 4 \times 2 = \fbox{32}$

The expression to be evaluated is: $(\sqrt{3}+\tan(5)^\circ)(\sqrt{3}+\tan(10)^\circ)(\sqrt{3}+\tan(15)^\circ)(\sqrt{3}+\tan(20)^\circ)(\sqrt{3}+\tan(25)^\circ)$

Notice that 25+5=30, so multiplying the first and last term might give us something.

$(\sqrt{3}+\tan(5)^\circ)(\sqrt{3}+\tan(25)^\circ)$ $=3+\sqrt{3}(\tan(5)^\circ+\tan(25)^\circ)+\tan(5)^\circ \tan(25)^\circ \, (*)$

We have,

$\displaystyle \tan(25+5)^\circ=\frac{\tan(5)^\circ+\tan(25)^\circ}{1-\tan(5)^\circ\tan(25)^\circ}=\frac{1}{\sqrt{3}}$

Rearranging, $\sqrt{3}(\tan(5)^\circ+\tan(25)^\circ)=1-\tan(5)^\circ\tan(25)^\circ$

Substituting in $(*)$ , the product of the first and last terms is 4. Similarly, it can be shown that the product of second and fourth term is 4.

Now we need the value $\tan(15)^\circ$ . This can be easily calculated using the double angle formula for tan. Hence

$\tan(30)^\circ=\frac{2\tan(15)^\circ}{1-\tan^2(15)^\circ}$

This gives us a quadratic in $\tan(15)^\circ$ . Solving and neglecting the negative root (because $\tan(\theta) > 0 \, \forall \, 0 \leq \theta <\frac{\pi}{2}$ ), we have

$\tan(15)^\circ=-\sqrt{3}+2$

Hence, the final answer is, $4\cdot 4 \cdot (\sqrt{3}-\sqrt{3}+2)=\fbox{32}$

We need to find a way to represent $\sqrt3 + \tan(5i^{\circ})$ as a fraction, so that terms may cancel when we take the given product. Consider the function $f(x) = \sqrt3 + \tan x,$ with all angles in degrees. We can write this function as $\begin{aligned} f(x) &= \sqrt 3 + \dfrac{\sin x}{\cos x} \\ &= \dfrac{\sqrt 3 \cos x + \sin x}{\cos x} \\ &= \dfrac{2\left(\dfrac{\sqrt 3}{2}\cos x + \dfrac12\sin x\right)}{\cos x} \\ &= \dfrac{2 \left(\cos 30^\circ \cos x + \sin 30^\circ \sin x\right)}{\cos x}. \end{aligned}$

Notice that we can use the addition formulas for cosine to rewrite the numerator of this fraction in a nicer way: we get $\qquad \sqrt3 + \tan x = \dfrac{2 \cos(30-x)}{\cos x} \qquad \textbf{(1)}$ as our result.

Applying equation $(1)$ to each term in our given product, we have $\begin{aligned}&\prod_{i =1}^5 [\sqrt 3 + \tan(5i^\circ)] \\ &= [\sqrt 3 + \tan 5^\circ][\sqrt 3 + \tan 10^\circ]\cdots[\sqrt 3 + \tan 25^\circ] \\ &= \frac{2 \cos(30-5)^\circ}{\cos 5^\circ} \frac{2 \cos(30-10)^\circ}{\cos 10^\circ} \cdots \frac{2 \cos(30-25)^\circ}{\cos 25^\circ} \\ &= \frac{2^5 \cos 5^\circ \cos 10^\circ \cos 15^\circ \cos 20^\circ \cos 25^\circ}{\cos 5^\circ \cos 10^\circ \cos 15^\circ \cos 20^\circ \cos 25^\circ}. \end{aligned}$ All of the cosines cancel, leaving only $2^5 = \boxed{32}.$

We know that $\cot(30)^{\circ}=\sqrt3$ . First, we see that $\tan(30)^{\circ}=\dfrac{\tan(i)^{\circ}+\tan(30-i)^{\circ}}{1-\tan(i)^{\circ}\tan(30-i)^{\circ}}.$ Equivalently $\cot(30)^{\circ}.(\tan(i)^{\circ}+\tan(30-i)^{\circ})=1-\tan(i)^{\circ}\tan(30-i)^{\circ}.$ So we have $(\cot(30)^\circ+\tan(30-i)^{\circ})(\cot(30)^\circ+\tan(i)^{\circ})=1+\cot^2(30)^{\circ}=4.$ Then $(\cot 30^{\circ}+\tan 5^{\circ})(\cot 30^{\circ}+\tan 25^{\circ})=4$ $(\cot 30^{\circ}+\tan 10^{\circ})(\cot 30^{\circ}+\tan 20^{\circ})=4$ $(\cot 30^{\circ}+\tan 15^{\circ})^2=4$ Since $\cot 30^{\circ}$ and $\tan 15^{\circ}$ is greater than zero, then $(\cot 30^{\circ}+\tan 15^{\circ})=2.$ So $\prod_{i=1}^{5} \left [ \sqrt{3}+\tan(5i)^{\circ} \right ]=2.4.4=32$

Each term can be written as

$2*[\frac{\frac{\sqrt{3}}{2}cos(5i^{0})+\frac{1}{2}sin(5i^{0}}{cos(5i^{0}))}]$

$\Rightarrow2*[\frac{sin((60+5i)^{0})}{cos(5i^{0}))}]$ {using $sin(A+B)$ }

Expanding the product, we can clearly see that the trgonometric terms of $sin$ and $cos$ in numerator and denominator cancel {using identity $sin(90^{0}-\theta)=cos(\theta)$ } only giving us $2^{5}=\boxed{32}$

We know that $\sqrt{3}=\tan{60}^\circ$ , so our product is: $\prod_{i=1}^5 [\tan{60}+\tan{5i}]=$ $(\tan{60}+\tan{5})(\tan{60}+\tan{10})\ldots(\tan{60}+\tan{25})$ From the tangent addition formula, $\displaystyle \tan (a+b)=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$ $\displaystyle \implies \tan{a}+\tan{b}=(\tan (a+b))(1-\tan{a}\tan{b})$

Applying this, we get that $(\tan{60}+\tan{5})(\tan{60}+\tan{10})\ldots(\tan{60}+\tan{25})=$ $(\tan{65})(1-\tan{60}\tan{5})\ldots (\tan{85})(1-\tan{60}\tan{25})=$ $(\tan{65})\ldots(\tan{85})(1-\tan{60}\tan{5})\ldots(1-\tan{60}\tan{25})=$ Since $\tan60=\sqrt{3}$ , $(\tan{65})\ldots(\tan{85})(1-\sqrt{3}\tan{5})\ldots(1-\sqrt{3}\tan{25})$ Now we arrange the binomials like such: $(1-\sqrt{3}\tan{5})(1-\sqrt{3}\tan{25})(1-\sqrt{3}\tan{10})(1-\sqrt{3}\tan{20})$ $(1-\sqrt{3}\tan15)=$ And multiply the first two pairs to get: $(1-\sqrt{3}(\tan5+\tan25)+3\tan5\tan25)(1-\sqrt{3}(\tan10+\tan20)+3\tan10\tan20)$ $(1-\sqrt{3}\tan15)=$ Applying our identity from the tangent addition formula to $\tan a+\tan b$ , we get: $(1-\sqrt{3}(\tan{30}(1-\tan5\tan25))+3\tan5\tan25)$ $(1-\sqrt{3}(\tan{30}(1-\tan10\tan20))+3\tan10\tan20)$ $(1-\sqrt{3}\tan15)=$ Simplifying and using the fact that $\displaystyle \frac{1}{\sqrt{3}}$ , $(4\tan5\tan25)(4\tan10\tan20)(1-\sqrt{3}\tan15)$ Putting this back into our original expression, we get: $16\tan5\tan10\tan20\tan25\tan65\ldots\tan85$ $(1-\sqrt{3}\tan15)=$ $16\tan{75}(1-\sqrt{3}\tan15)$ Everything $\tan a$ cancels except for $\tan75$ . Simplifying, we get $16\tan75-\sqrt{3}=\frac{16\sin75-\sqrt{3}\cos75}{\cos75}=$ $\frac{32(\frac{1}{2}\sin75-\frac{\sqrt{3}}{2}\cos75)}{\cos75}=\frac{32\sin15}{\cos75}=\boxed{32}$ Where the sine addition formula was applied backwards in the above line.

find the identity of expression

$\sqrt{3}+\tan\theta=\frac{1}{\cos\theta}(\sqrt{3}\cos\theta+\sin\theta)$ $=\frac{2}{\cos\theta}(\sin\frac{\pi}{3}+\cos\theta\sin\theta)=\frac{2}{\cos\theta}\cdot\sin(\theta+\frac{\pi}{3})$

when you write all term of this problem's expression you will find that expression of $\sin,\cos$ cancle each other and you will get $2^5$ is answer

$\tan 30^\circ=\tan 15^\circ$ , $\frac{1}{\sqrt3}=\frac{2 \tan 15^\circ}{1-\tan^2 15^\circ}$ Solving this we get $\tan 15^\circ=-\sqrt 3 +2$ ( $-\sqrt 3 - 2$ is ignored since $\tan 15^\circ$ is positive) For the rest $( \sqrt 3 + \tan 5^\circ )( \sqrt 3 + \tan 25^\circ )( \sqrt 3 + \tan 10^\circ )( \sqrt 3 + \tan 20^\circ )$ $(3 + (\sqrt 3)(\tan 5^\circ + \tan 25^\circ) + \tan 5^\circ \tan 25^\circ)( ... )$ From $\tan 30^\circ=\tan (5^\circ + 25^\circ)=\tan (10^\circ + 20^\circ )$ , $\frac {1}{ \sqrt 3 }=\frac { \tan 5^\circ + \tan 25^\circ }{ 1 - (\tan 5^\circ)(\tan 25^\circ)} = ...$ , $1= (\sqrt 3)(\tan 5^\circ + \tan 25^\circ) + \tan 5^\circ \tan 25^\circ = ...$ So in the end we will have $( 3 + 1 )( 3+1)( \sqrt 3 - \sqrt 3 + 2 ) = 32$

Remembering that:

• $tan(15°) = 2 - \sqrt{3}$

• $tan(30°) = \frac{1}{\sqrt{3}}$

• $tan(a + b) = \frac {tan(a) + tan(b)}{1 - tan(a) tan(b)} \iff tan(a) tan(b) = 1 - \frac{tan(a) + tan(b)} {tan(a+ b)}$

We have, for $0° \lt x \lt 15°$ : $(\sqrt{3} + tan(15° + x)) (\sqrt{3} + tan(15° - x)) =$ $3 + \sqrt{3} tan(15° + x) + \sqrt{3} tan(15° - x) + tan(15° + x) tan(15° -x) =$ $3 + \sqrt{3} tan(15° + x) + \sqrt{3} tan(15° - x) + 1 - \frac {tan(15° + x)}{tan(30°)} - \frac {tan(15° - x)}{tan(30°)} =$ $= 4$

Now we can rewrite the original expression as: $\prod_{i = 1}^{5} (\sqrt{3} + tan(5i°) ) = \prod_{i = -2}^{2} (\sqrt{3} + tan(15° + 5i°) ) =$

$(\sqrt{3} + tan(15°) \prod_{i = 1}^{2} ((\sqrt{3} + tan(15° + 5i°)) (\sqrt{3} + tan(15° - 5i°))) =$

$= 2 \times 4 \times 4= 32$