(i) \( \frac{1}{\tan x}+\tan x=\frac{\tan x}{\sin x} \)

Solve the equation \( \frac{1}{\tan x}+\tan x=\frac{\tan x}{\sin x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{1}{\tan\left(x\right)}+\tan\left(x\right)=\frac{\tan\left(x\right)}{\sin\left(x\right)}\) - step1: Find the domain: \(\frac{1}{\tan\left(x\right)}+\tan\left(x\right)=\frac{\tan\left(x\right)}{\sin\left(x\right)},x\neq \frac{k\pi }{2},k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\frac{1}{\frac{\sin\left(x\right)}{\cos\left(x\right)}}+\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}}{\sin\left(x\right)}\) - step3: Simplify the expression: \(\frac{\cos\left(x\right)}{\sin\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}}{\sin\left(x\right)}\) - step4: Simplify the expression: \(\frac{\cos\left(x\right)}{\sin\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)\sin\left(x\right)}\) - step5: Reduce the fraction: \(\frac{\cos\left(x\right)}{\sin\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{1}{\cos\left(x\right)}\) - step6: Multiply both sides of the equation by LCD: \(\left(\frac{\cos\left(x\right)}{\sin\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)\cos\left(x\right)\sin\left(x\right)=\frac{1}{\cos\left(x\right)}\times \cos\left(x\right)\sin\left(x\right)\) - step7: Simplify the equation: \(\cos^{2}\left(x\right)+\sin^{2}\left(x\right)=\sin\left(x\right)\) - step8: Rewrite the expression: \(1=\sin\left(x\right)\) - step9: Swap the sides: \(\sin\left(x\right)=1\) - step10: Use the inverse trigonometric function: \(x=\arcsin\left(1\right)\) - step11: Calculate: \(x=\frac{\pi }{2}\) - step12: Add the period: \(x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\) - step13: Check if the solution is in the defined range: \(x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z},x\neq \frac{k\pi }{2},k \in \mathbb{Z}\) - step14: Find the intersection: \(x \in \varnothing \) The equation \( \frac{1}{\tan x}+\tan x=\frac{\tan x}{\sin x} \) has no solution.