Question

If $\tan 5 A =\tan \theta$ and $\tan (3 A + B )=1$, and $QA = B +\theta-\pi / p, n \in Z$. Also it is given that the value of $\sqrt{ P ^{3}+ R \cdot Q ^{2}}=10$, then find the value of $R$

Answer 1

$\tan 5 A =\tan \theta $
$\Rightarrow 5 A = n \pi+\theta$
$\Rightarrow A =\frac{ n \pi}{5}+\frac{\theta}{5}$
$\tan (3 A + B )=1 $
$\Rightarrow \tan (3 A + B )=\tan \frac{\pi}{4}$
$\Rightarrow 3 A + B = n \pi+\frac{\pi}{4}$
$\Rightarrow B = n \pi-\frac{3 n \pi}{5}-\frac{3 \theta}{5}+\frac{\pi}{4}=\frac{2 n \pi}{5}-\frac{3 \theta}{5}+\frac{\pi}{4}$
By the equation $QA = B +\theta-\frac{\pi}{ p }( n \in Z )$
$\Rightarrow \frac{ Qn \pi}{5}+\frac{ Q \theta}{5}=\frac{2 n \pi}{5}-\frac{3 \theta}{5}+\frac{\pi}{4}+\theta-\frac{\pi}{ p }$
$\Rightarrow \frac{ Qn \pi}{5}+\frac{ Q \theta}{5}=\frac{2 n \pi}{5}+\frac{2 \theta}{5}+\frac{\pi}{4}-\frac{\pi}{ p }$
$\Rightarrow Q =2, p =4$,
Also $\sqrt{ p ^{3}+ R \cdot Q ^{2}}=10$
$\Rightarrow \sqrt{64+ R \cdot 4}=10$
$ \Rightarrow R =9$