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.
Best Answer
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 $$\cos\arctan\frac{3}{4}=\sqrt{1-\sin^2\arctan\frac{3}{4}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.$$