<s>[INST] If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$. 
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.
 [/INST] 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  $\sin t\cos t\ne0$
</s>