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If $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi}{2},$ then the value of $\left(\cos \frac{P}{3}-i \sin \frac{P}{3}\right)^{3}+(\cos Q+i \sin Q) \\ (\cos R-i \sin R)+(\cos P-i \sin P) \\ (\cos Q-i \sin Q)(\cos R-i \sin R) \text { is }$
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The correct answer is:
$-i$
Given that, $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi}{2}$
$\therefore \quad Q=R=\frac{\pi}{4}\left(\because P+Q+R=180^{\circ}\right)$
Now, $\left(\cos \frac{P}{3}-i \sin \frac{P}{3}\right)^{3}+(\cos Q+i \sin Q)$
$(\cos R-i \sin R)$
$+(\cos P-i \sin P)(\cos Q-i \sin Q)$
$\quad(\cos R-i \sin R)$
$=\left(e^{-i \frac{P}{3}}\right)^{3}+e^{i Q} \cdot e^{-i R}+e^{-i \rho} e^{-i Q} \cdot e^{-i R}$
$\left(\because \cos \theta+i \sin \theta=e^{i \theta}\right)$
$=e^{-i P}+e^{\gamma Q-R)}+e^{-i(P+Q+R)}$
$=e^{-i \pi / 2}+e^{(0)}+e^{-i \pi} \quad\left(Q=R=\frac{\pi}{2} P=\frac{\pi}{2}\right)$
$=-i+1+(-1)=-i$
$\therefore \quad Q=R=\frac{\pi}{4}\left(\because P+Q+R=180^{\circ}\right)$
Now, $\left(\cos \frac{P}{3}-i \sin \frac{P}{3}\right)^{3}+(\cos Q+i \sin Q)$
$(\cos R-i \sin R)$
$+(\cos P-i \sin P)(\cos Q-i \sin Q)$
$\quad(\cos R-i \sin R)$
$=\left(e^{-i \frac{P}{3}}\right)^{3}+e^{i Q} \cdot e^{-i R}+e^{-i \rho} e^{-i Q} \cdot e^{-i R}$
$\left(\because \cos \theta+i \sin \theta=e^{i \theta}\right)$
$=e^{-i P}+e^{\gamma Q-R)}+e^{-i(P+Q+R)}$
$=e^{-i \pi / 2}+e^{(0)}+e^{-i \pi} \quad\left(Q=R=\frac{\pi}{2} P=\frac{\pi}{2}\right)$
$=-i+1+(-1)=-i$
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