c) \( \cos ^{2} x+\cos ^{2} x \cdot \tan ^{2} x=1 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\cos^{2}\left(x\right)+\cos^{2}\left(x\right)\tan^{2}\left(x\right)=1\) - step1: Find the domain: \(\cos^{2}\left(x\right)+\cos^{2}\left(x\right)\tan^{2}\left(x\right)=1,x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Multiply the terms: \(\cos^{2}\left(x\right)+\left(\cos\left(x\right)\tan\left(x\right)\right)^{2}=1\) - step3: Move the expression to the left side: \(\cos^{2}\left(x\right)+\left(\cos\left(x\right)\tan\left(x\right)\right)^{2}-1=0\) - step4: Calculate: \(\cos^{2}\left(x\right)+\cos^{2}\left(x\right)\tan^{2}\left(x\right)-1=0\) - step5: Rewrite the expression: \(\cos^{2}\left(x\right)+\cos^{2}\left(x\right)\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)^{2}-1=0\) - step6: Simplify: \(0=0\) - step7: The statement is true: \(x \in \mathbb{R}\) - step8: Check if the solution is in the defined range: \(x \in \mathbb{R},x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step9: Find the intersection: \(x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) To solve the equation \[ \cos^{2} x + \cos^{2} x \cdot \tan^{2} x = 1, \] we can start by using the identity for \(\tan x\): \[ \tan x = \frac{\sin x}{\cos x}. \] Thus, we can rewrite \(\tan^{2} x\) as: \[ \tan^{2} x = \frac{\sin^{2} x}{\cos^{2} x}. \] Substituting this into the equation gives: \[ \cos^{2} x + \cos^{2} x \cdot \frac{\sin^{2} x}{\cos^{2} x} = 1. \] This simplifies to: \[ \cos^{2} x + \sin^{2} x = 1. \] Using the Pythagorean identity \(\sin^{2} x + \cos^{2} x = 1\), we see that this equation is always true for all \(x\) where \(\cos x \neq 0\). The only restriction comes from the fact that \(\tan x\) is undefined when \(\cos x = 0\), which occurs at: \[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}. \] Thus, the solution to the equation is: \[ x \neq \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}. \] This means \(x\) can take any real value except for those specific points where \(\cos x\) is zero.