समाधान {l}{sin(A)=8/17}{a=tan(A)+sec(A)}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { \sin(A) = \frac{8}{17} }\\ { a = \tan(A) + \sec(A) }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

A, a, b को लागि हल गर्नुहोस् (complex solution)

$$A=2\pi n_{1}+\arctan(\frac{8}{15})\text{, }n_{1}\in \mathrm{Z}\text{, }a=\frac{5}{3}\approx 1.666666667\text{, }b=\frac{5}{3}\approx 1.666666667$$

$$A=2\pi n_{2}+\pi -\arctan(\frac{8}{15})\text{, }n_{2}\in \mathrm{Z}\text{, }a=-\frac{5}{3}\approx -1.666666667\text{, }b=-\frac{5}{3}\approx -1.666666667$$

A, a, b को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}A=2\pi n_{4}+\arcsin(\frac{8}{17})\text{, }n_{4}\in \mathrm{Z}\text{, }a=\frac{5}{3}\approx 1.666666667\text{, }b=\frac{5}{3}\approx 1.666666667\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arcsin(\frac{8}{17})=\frac{\pi \left(2n_{1}+1\right)}{2}\right)\right)\\A=2\pi n_{5}+\pi -\arcsin(\frac{8}{17})\text{, }n_{5}\in \mathrm{Z}\text{, }a=-\frac{5}{3}\approx -1.666666667\text{, }b=-\frac{5}{3}\approx -1.666666667\text{, }&\exists n_{3}\in \mathrm{Z}\text{ : }\left(\exists n_{3}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+\pi -\arcsin(\frac{8}{17})=\frac{\pi \left(2n_{1}+1\right)}{2}\right)\right)\end{matrix}\right.$$