Finding $\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right) $  assume $\displaystyle \frac{\pi}{14} = \theta$ then $\tan (14\theta) = 0$  wan,t be able to go further, some help me
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(1) Evaluation of $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = $ (2) Evaluation of $\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right) = $  $\bf{My\; Try::(1):}$ Let $\displaystyle \frac{\pi}{7} = \theta\Rightarrow \pi = 7\theta\Rightarrow (\pi-3\theta) = 4\theta\Rightarrow \tan(\pi-3\theta) = \tan(4\theta)$ So $\displaystyle \tan (3\theta) = -\tan (4\theta)\Rightarrow \frac{3\tan \theta - \tan^3\theta}{1-3\tan^2 \theta} = -\left(\frac{2\tan 2\theta}{1-\tan^2 2\theta}\right) = \frac{4\tan^3 \theta-4\tan \theta}{1+\tan^4 \theta-6\tan^2 \theta}$ So $\displaystyle \left(\frac{3-\tan^2 \theta}{1-3\tan^2 \theta}\right) = \left(\frac{4\tan^2 \theta-4}{1+\tan^4 \theta-6\tan^2 \theta}\right)\Rightarrow \tan^6\theta-21\tan^4 \theta+21\tan^2 \theta-7=0$ Now Let $\displaystyle \tan \theta = y\;,$ then eqn. convert into $\displaystyle y^6-21y^4+21y^2-7=0$ and equation has a roots $\displaystyle y = \tan \left(\frac{\pi}{7}\right)\;,\tan \left(\frac{2\pi}{7}\right)\;,\tan \left(\frac{3\pi}{7}\right)\;,\tan \left(\frac{4\pi}{7}\right)\;,\tan \left(\frac{5\pi}{7}\right)\;,\tan \left(\frac{6\pi}{7}\right)\;$. So Product of Roots is $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right)\cdot \tan \left(\frac{4\pi}{7}\right)\cdot \tan \left(\frac{5\pi}{7}\right)\cdot \tan \left(\frac{6\pi}{7}\right) = -7$ Now $\displaystyle \tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi-\frac{3\pi}{7}\right)=-\tan\left(\frac{3\pi}{7}\right)$ Similarly $\displaystyle \tan\left(\frac{5\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)$ and $\displaystyle \tan\left(\frac{6\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)$ So $\displaystyle \tan\left(\frac{\pi}{7}\right)\cdot \tan\left(\frac{2\pi}{7}\right)\cdot \tan\left(\frac{3\pi}{7}\right)=\sqrt{7}$ I did not understand how can i calculate $(2)$ one using the same method:, bcz it is very lengthy can we solve the $(1)$ and $(2)$ question any other method. If yes then plz explain me, Thanks :: Thanks
33077
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$  I've let $u=x^2+3$ but can't seem to get the right answer.  Really not sure what to do.
69384
By looking through an book, I found this interesting series  To prove that:  $$\tan(\theta)+\tan \left(\theta+ \frac{\pi}{n} \right) + \tan\left(\theta + \frac{2\pi}{n}\right) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$  I have tried the Gaussian pairing trick and using:  $$\tan A + \tan B = \tan(A+B) (1-\tan A \tan B)$$  And  $$\tan A \tan B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B)+\cos(A+B)}$$  However I could not do anything of great use.  I also considered perhaps roots of unity of:  $$z^{2n} = \cos \theta + i\sin \theta$$  But that grew to no real use, but it may be an idea for others.  I know I can sum an arithmetic series of the arguments for the cosine and sine functions but I cannot find a source on the internet about the tangent function.   Thank You.
120372
If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$   My 1st  approach :   $\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $  $\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.   My IInd approach :   U\sing Euler substitution :   \since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)  Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$  and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$   By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$  $$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$  Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
98400
Are $\cos^2 \theta$ and $\cos \theta^2$ the same? I mean be it $\sin,\cos, \tan ,\cot ,\sec,\csc$.  Are they same? Please help a maths noob here.
84821
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$   I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
24894
Show that,   $$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$  There is a mixed of sin and tan, how  can I simplify this to $\frac{\pi}{4}$  We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
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If $\alpha$ and $\beta$ are the solution of $$a \cdot \tan\theta + b \cdot \sec\theta = c$$, then show that $$ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$$  I did the following:     $$a \cdot \tan\theta + b \cdot \sec\theta = c$$       Squaring both sides,  $$a^2 \cdot \tan^2\theta + b^2 \cdot \sec \theta + 2 \cdot a \cdot b \cdot \tan \theta \cdot \sec \theta = c^2$$      This implies $$(a^2 + b^2) \cdot \sin^2 \theta + 2 \cdot a \cdot b \cdot \sin \theta + (b^2 - c^2) = 0$$       Hence I get $$\sin \alpha + \sin \beta = \frac{-2ab}{a^2 + c^2}$$ and $$\sin \alpha \cdot \sin \beta = \frac{b^2 - c^2}{a^2 - c^2}$$   But I am not able to solve it further.
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Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$    \begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&amp;=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &amp;=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align}  I am stuck here.
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$$\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$$  I tried substituting $x=z^2$ ... also $x=\tan^2 \theta$ ... but couldn't solve it either ways... if someone can help then it would be good.
10274
Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
53559
I'm having problem with showing that: $$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$  I would need some help in the right direction
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Problem:   Find the value of $$\sum_{r=0}^{44} \tan^2(2r+1)$$  Note: The angles here are in degrees.    I don't know how to solve this question because trigonometric simplifications didn't get me anywhere. I think there was a method to solve this question using complex numbers which I no longer remember. Any hint/help will be appreciated.
33229
Find the value of $$\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+......+\cos^2(\theta+179^{\circ})$$  Can anyone teach me where to start with? I've no idea.
25616
Find the minimum value of      $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$   $a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\ c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $  $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\  =\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\ \color{blue}{\text{By using the AM-GM inequlity}} \\ \color{blue}{x+\dfrac{1}{x} \geq 2} \\  =2+2+2=6 $  Which is not in options.  But I am not sure if I can use that $ AM-GM$ inequality in this case.  I look for a short and simple way .  I have studied maths upnto $12$th grade .
52447
Prove the following:     $\frac{\sin(a)}{\sin(b)} &lt; \frac{a}{b} &lt; \frac{\tan(a)}{\tan(b)}$ where $0 &lt; b &lt; a &lt; \frac{\pi}{2}$   Hello everyone, I am trying to create some sort of function or maybe see an application of the mean value theorem, but I am just not getting it. Any help would be greatly appreciated!
21305
How to simplify $$\arctan \left(\frac{1}{2}\tan (2A)\right) +  \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) $$ for  $0&lt; A&lt; \pi /4$?  This is one of the problems in a book I'm using. It is actually an objective question , with 4 options given , so i just put $A=\pi /4$ (even though technically its disallowed as $0&lt; A&lt; \pi /4$) and got the answer as $\pi $ which was one of the options , so that must be the answer (and it is weirdly written in options as $4 \arctan (1) $ ).  Still , I'm not able to actually solve this problem. I know the formula for sum of three arctans , but it gets just too messy and looks hard to simplify and it is not obvious that the answer will be constant for all $0&lt; A&lt; \pi /4$. And I don't know of any other way to approach such problems.