Observe the following columns:

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Professor PC Thomas Classes · Answer

$ \begin{aligned} & 	ext { (A) } \because \sin x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}ight), \sin \left(2 \pi-\frac{\pi}{6}ight) \ & 	herefore x=\frac{7 \pi}{6}, \frac{11 \pi}{6} \ & 	ext { and } \cos x=-\frac{\sqrt{3}}{2}=-\cos \frac{\pi}{6}=\cos \left(\pi-\frac{\pi}{6}ight), \cos \left(\pi+\frac{\pi}{6}ight) \ & 	herefore x=\frac{5 \pi}{6}, \frac{7 \pi}{6} \end{aligned} $ \ From Equations (i) and (ii). It is clear that $ \begin{aligned} & \alpha=\frac{7 \pi}{6}, \beta \frac{11 \pi}{6}, \gamma=\frac{5 \pi}{6} \Rightarrow \alpha+\beta=3 \pi(S), \beta-\gamma=\pi(Q) \ & 	herefore \cot x=-\sqrt{3}=-\cot \frac{\pi}{6}=\cot \left(\pi-\frac{\pi}{6}ight), \cot \left(2 \pi-\frac{\pi}{6}ight) \ & 	herefore x=\frac{5 \pi}{6}, \frac{11 \pi}{6} 	ext { and } \operatorname{cosec} x=-2=-\operatorname{cosec} \frac{\pi}{6}=\operatorname{cosec}\left(\pi+\frac{\pi}{6}ight), \operatorname{cosec}\left(2 \pi-\frac{\pi}{6}ight) \ & x=\frac{7 \pi}{6}, \frac{11 \pi}{6}(C) \because \sin x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}ight), \sin \left(2 \pi-\frac{\pi}{6}ight) \ & 	herefore x=\frac{7 \pi}{6}, \frac{11 \pi}{6} \end{aligned} $ $ \begin{aligned} 	an x & =\frac{1}{\sqrt{3}}=	an \frac{\pi}{6}=	an \frac{\pi}{6}, 	an \left(\pi+\frac{\pi}{6}ight) \ 	herefore \quad x & =\frac{\pi}{6}, \frac{7 \pi}{6} \end{aligned} $ \ From Eqs. (i) and (ii), it is clear that $$ \begin{aligned} & \alpha=\frac{7 \pi}{6}, \beta=\frac{11 \pi}{6}, \gamma=\frac{\pi}{6} \ & \Rightarrow \alpha+\beta=3 \pi,(S), \beta+\gamma=2 \pi(T), \alpha-\gamma=\pi(R) \end{aligned} $$ From Eqs. (i) and (ii), it is clear that $$ \begin{aligned} & \alpha=\frac{11 \pi}{6}, \beta=\frac{5 \pi}{6}, \gamma=\frac{7 \pi}{6} \ & \Rightarrow \beta+\gamma=2 \pi(T), \alpha-\beta=\pi(P) \end{aligned} $$