Find the coordinates of the point C which lies on the line AB produced such that $\displaystyle \mathrm{AC}=2 \mathrm{BC}$, where coordinates of points A and B are $\displaystyle (-1, 7)$ and $\displaystyle (4, -3)$ respectively.

Let coordinates of point C be ( \(\displaystyle \mathrm{x}, \mathrm{y}\) ) \(\displaystyle \mathrm{AC}=2 \mathrm{BC}\) ⇒ B is mid-point of AC \(\displaystyle \Rightarrow \frac{-1+\mathrm{x}}{2}=4 \Rightarrow \mathrm{x}=9\) \(\displaystyle \frac{7+\mathrm{y}}{2}=-3 \Rightarrow \mathrm{y}=-13\) ∴ Coordinates of C are \(\displaystyle (9,-13)\) \[\begin{aligned} & (\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=x+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A} \\ & \Rightarrow \sin ^{2} \mathrm{~A}+\operatorname{cosec}^{2} \mathrm{~A}+2+\cos ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}+2=x+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A} \\ & \Rightarrow 1+2+2+1+\cot ^{2} \mathrm{~A}+1+\tan ^{2} \mathrm{~A}=x+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A} \\ & \therefore x=7 \end{aligned} \]\[\begin{aligned} & \frac{3 \sin 30^{\circ}-4 \sin ^{3} 30^{\circ}}{2 \sin ^{2} 50^{\circ}+2 \cos ^{2} 50^{\circ}} \\ & =\frac{3 \times \dfrac{1}{2}-4 \times \dfrac{1}{8}}{2\left(\sin ^{2} 50^{\circ}+\cos ^{2} 50^{\circ}\right)} \\ & =\frac{\dfrac{3}{2}-\dfrac{1}{2}}{2 \times 1} \\ & =\frac{1}{2} \end{aligned} \]