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If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.

My approach:- $$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$ By using this, we get value of $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.

I want know if there's any way to solve this problem without calculator.

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  • $\begingroup$ Is this $$0<\theta<\frac{\pi}{18}$$? $\endgroup$ – Dr. Sonnhard Graubner Apr 14 at 8:21
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Consider the right triangle with sides $3,4,5$, whose angle opposite the side $3$ is $9\theta$. Then: $$\sin 9\theta =\frac35 \Rightarrow 5\sin 9\theta=3; \\ \cos 9\theta =\frac45 \Rightarrow 5\cos 9\theta =4;\\ 3\csc 3\theta - 4\sec 3\theta=\frac{3}{\sin 3\theta}-\frac4{\cos 3\theta}=\frac{3\cot 3\theta-4\sin 3\theta}{\sin 3\theta \cos 3\theta}=\\ \frac{5\sin 9\theta \cos3\theta-5\cos 9\theta \sin 3\theta}{\sin 3\theta \cos 3\theta}=\\ \frac{5\sin (9\theta -3\theta)}{0.5\sin 6\theta}=10.$$

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  • $\begingroup$ Edit your answer with cos instead of cot, and your approach is very simple and easy. $\endgroup$ – Abhishek Kumar Apr 17 at 18:20
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    $\begingroup$ Fixed the typo, thanks. $\endgroup$ – farruhota Apr 17 at 18:23
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We have $$\theta=\frac{1}{9}\arctan\frac{3}{4}.$$

Thus, $$\frac{3}{\sin3\theta}-\frac{4}{\cos3\theta}=\frac{3}{\sin\frac{1}{3}\arctan\frac{3}{4}}-\frac{4}{\cos\frac{1}{3}\arctan\frac{3}{4}}=$$ $$=\frac{3\cos\frac{1}{3}\arctan\frac{3}{4}-4\sin\frac{1}{3}\arctan\frac{3}{4}}{\sin\frac{1}{3}\arctan\frac{3}{4}\cos\frac{1}{3}\arctan\frac{3}{4}}=$$ $$=\frac{10\left(\frac{3}{5}\cos\frac{1}{3}\arctan\frac{3}{4}-\frac{4}{5}\sin\frac{1}{3}\arctan\frac{3}{4}\right)}{\sin\frac{2}{3}\arctan\frac{3}{4}}=$$ $$=\frac{10\sin\left(\arctan\frac{3}{4}-\frac{1}{3}\arctan\frac{3}{4}\right)}{\sin\frac{2}{3}\arctan\frac{3}{4}}=10.$$ I used $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta,$$ $$\sin\arctan\frac{3}{4}=\sqrt{1-\cos^2\arctan\frac{3}{4}}=$$ $$=\sqrt{1-\frac{1}{1+\tan^2\arctan\frac{3}{4}}}=\sqrt{1-\frac{1}{1+\left(\frac{3}{4}\right)^2}}=\frac{3}{5}$$ and $$\cos\arctan\frac{3}{4}=\sqrt{1-\sin^2\arctan\frac{3}{4}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.$$

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  • $\begingroup$ How did you get from the fourth line to the fifth line? $\endgroup$ – Toby Mak Apr 17 at 9:12
  • $\begingroup$ @Toby Mak I added something. See now. $\endgroup$ – Michael Rozenberg Apr 17 at 11:00
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    $\begingroup$ It makes sense now. Thanks! $\endgroup$ – Toby Mak Apr 17 at 11:34
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Hint: We get $$\theta=\frac{1}{9}\arctan(\frac{3}{4})$$ so we get $$3\csc\left(\frac{1}{3}\arctan(\frac{3}{4})\right)-4\sec\left(\frac{1}{3}\arctan(\frac{3}{4})\right)=10$$

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  • $\begingroup$ Graubner, Can you please last line? $\endgroup$ – lab bhattacharjee Apr 14 at 10:01
  • $\begingroup$ To the up-voter, how do you explain the last line? $\endgroup$ – lab bhattacharjee Apr 14 at 10:01
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    $\begingroup$ Explain your solution plz $\endgroup$ – Abhishek Kumar Apr 14 at 11:17
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Set $3\theta=t$

$\tan3t=\dfrac34\implies\dfrac{\sin3t}3=\dfrac{\cos3t}4=\dfrac1k$

$\implies(k\sin3t)^2+(k\cos3t)^2=3^2+4^2\implies k=5$ as $0<t<\dfrac\pi6$

$$F(t)=\dfrac3{\sin t}-\dfrac4{\cos t}=\dfrac{k\sin3t}{\sin t}-\dfrac{k\cos3t}{\cos t}$$

Method$\#1:$

As $\sin t\cos t\ne0,$ using $\sin3t,\cos3t$ formula,

$$F(t)=k(3-4\sin^2t)-k(4\cos^2t-3)=k(3+3-4)$$

Method$\#2:$

$$F(t)=\dfrac{k\sin(3t-t)}{\dfrac{\sin2t}2}=2k$$ as $\sin t\cos t\ne0$

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  • $\begingroup$ @AbhishekKumar, Could you follow the method here? $\endgroup$ – lab bhattacharjee Apr 16 at 6:17

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