c) $$ \cos ^{2} x+\cos ^{2} x \cdot an ^{2} x=1 $$

Question

c) $$ \cos ^{2} x+\cos ^{2} x \cdot 	an ^{2} x=1 $$

UpStudy AI · Accepted Answer

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.
The equation $$\cos^{2} x + \cos^{2} x \cdot \tan^{2} x = 1$$ is always true for all $$x$$ except when $$\cos x = 0$$, which happens at $$x = \frac{\pi}{2} + k\pi$$ for any integer $$k$$.

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

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