Solve tanA+sinA/tanA-sinA=pi/b

$$\frac { \tan A + \sin A } { \tan A - \sin A } = \pi / b$$

Solve for b

$$b=-\pi \left(\tan(\frac{A}{2})\right)^{2}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(A>\frac{\pi n_{1}}{2}\text{ and }A<\frac{\pi n_{1}}{2}+\frac{\pi }{2}\right)$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by b.

$$bCosI(A)\left(SinI(A)\left(1-CosI(A)\right)\right)^{-1}\left(\tan(A)+\sin(A)\right)=\pi$$

Use the distributive property to multiply SinI(A) by 1-CosI(A).

$$bCosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\left(\tan(A)+\sin(A)\right)=\pi$$

Use the distributive property to multiply bCosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1} by \tan(A)+\sin(A).

$$bCosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+bCosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A)=\pi$$

Combine all terms containing b.

$$\left(CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A)\right)b=\pi$$

The equation is in standard form.

$$\frac{\cos(A)\left(\sin(A)+\tan(A)\right)}{\sin(A)\left(-\cos(A)+1\right)}b=\pi$$

Divide both sides by CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A).

$$\frac{\frac{\cos(A)\left(\sin(A)+\tan(A)\right)}{\sin(A)\left(-\cos(A)+1\right)}b\sin(A)\left(-\cos(A)+1\right)}{\cos(A)\left(\sin(A)+\tan(A)\right)}=\frac{\pi \sin(A)\left(-\cos(A)+1\right)}{\cos(A)\left(\sin(A)+\tan(A)\right)}$$

Dividing by CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A) undoes the multiplication by CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A).

$$b=\frac{\pi \sin(A)\left(-\cos(A)+1\right)}{\cos(A)\left(\sin(A)+\tan(A)\right)}$$

Divide \pi by CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\tan(A)+CosI(A)\left(SinI(A)-SinI(A)CosI(A)\right)^{-1}\sin(A).

$$b=\frac{\pi \left(-\cos(A)+1\right)}{\cos(A)+1}$$

Variable b cannot be equal to 0.

$$b=\frac{\pi \left(-\cos(A)+1\right)}{\cos(A)+1}\text{, }b\neq 0$$

Solve for A

$$A=2\pi n_{1}+\arccos(\frac{b-\pi }{b+\pi })-\pi \text{, }n_{1}\in \mathrm{Z}$$

$$A=2\pi n_{2}-\arccos(\frac{b-\pi }{b+\pi })+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }b>0$$