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My attempt : $$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$ Change into variable form $$\dfrac{(1+a^2)(1-b^2)}{ab}$$ $$\dfrac{1+a^2-b^2-a^2b^2}{ab}$$ I'm stuck here also I don't think this is the correct way.

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  • $\begingroup$ Use trigonometric identities. $\endgroup$
    – cjferes
    Jul 5, 2020 at 6:43
  • $\begingroup$ @cjferes $\sec^2\theta-\tan^2\theta=1$ use this identity, Am I right? $\endgroup$
    – Ken
    Jul 5, 2020 at 6:46
  • $\begingroup$ Hint: Write $$\tan x=\frac{\sin x}{\cos x}$$ and simplify. You've to use compound angle results of $\sin$ and $\cos$. Also, your mistake is that $\tan(π-x)=-\tan x$. $\endgroup$
    – SarGe
    Jul 5, 2020 at 6:53
  • $\begingroup$ Another way may be noticing that $\tan\frac{\pi}{12} = 2 - \sqrt3$ and $\tan\frac{5\pi}{12} = \cot\frac{\pi}{12} = 2 + \sqrt3$. To calculate $\tan\frac{\pi}{12}$ one may use $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ . $\endgroup$
    – IPPK
    Jul 5, 2020 at 7:14

2 Answers 2

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$$\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$$ $$=\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{5\pi}{12}}$$ $$=\dfrac{4}{\left(\dfrac{2\tan\frac{5\pi}{12}}{(1+\tan^2\frac{5\pi}{12})}\right)\left(\dfrac{2\tan\frac{\pi}{12}}{{1-\tan^2\frac{\pi}{12}}}\right)}$$ Use trig. identity, $\frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta$, $\frac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$, $$=\dfrac{4}{\left(\sin\left(2\frac{5\pi}{12}\right)\right)\left(\tan\left(2\frac{\pi}{12}\right)\right)}$$ $$=\dfrac{4}{\left(\sin\frac{5\pi}{6}\right)\left(\tan\frac{\pi}{6}\right)}$$ $$=\dfrac{4}{\left(\frac12\right)\left(\frac{1}{\sqrt 3}\right)}$$ $$=8\sqrt3$$

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  • $\begingroup$ Sorry, how does $1-\tan^2\frac{11\pi}{12}$ become $1+\tan^2\frac{\pi}{12}$ $\endgroup$
    – Ken
    Jul 5, 2020 at 7:06
  • $\begingroup$ The question is actually different if you see. The second bracket in Numerator has $-$ sign. $\endgroup$
    – Z Ahmed
    Jul 5, 2020 at 7:07
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    $\begingroup$ @HarishChandraRajpoot Am I misunderstood anything here $1-\tan^2\frac{11\pi}{12}=1-\tan^2\left(\pi-\frac{\pi}{12}\right)=1-(-\tan\frac{\pi}{12})(-\tan\frac{\pi}{12})=1-\tan^2\left(\frac{\pi}{12}\right)$ $\endgroup$
    – Ken
    Jul 5, 2020 at 7:15
  • $\begingroup$ I have the same question as @Ken asked. $\endgroup$ Jul 5, 2020 at 7:19
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    $\begingroup$ @Ken: You are very right. I just forgot there is square term. I fixed it thanks $\endgroup$ Jul 5, 2020 at 7:19
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We can write $$ F=\frac{(1+\tan^2(5\pi/12))(1-\tan^2(\pi/12))}{\tan(5\pi/12) \tan(\pi/12)}$$ $$\implies F=\frac{2}{\frac{2\tan(\pi/12)}{1-\tan^2(\pi/12)}}\frac{\sec^2(5\pi/12)}{\tan(5\pi/12)}.$$ $$F=2 \cot (\pi/6) \frac{\csc^2(\pi/12)}{\cot(\pi/12)}=4 \sqrt{3} \csc(\pi/6)=8\sqrt{3}$$

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