$$ (\frac{1- an ( heta )}{\cos ( heta )})^{2}=\frac{1-\sin ( heta )}{1+\sin ( heta )} $$

Question

$$ (\frac{1-	an (	heta )}{\cos (	heta )})^{2}=\frac{1-\sin (	heta )}{1+\sin (	heta )} $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\left(\frac{1-\tan\left(\theta \right)}{\frac{1}{\frac{1}{\tan\left(\theta \right)}}}\right)^{2}=\frac{1-\tan\left(\theta \right)}{1+\tan\left(\theta \right)}$$
- step1: Find the domain:
  $$\left(\frac{1-\tan\left(\theta \right)}{\frac{1}{\frac{1}{\tan\left(\theta \right)}}}\right)^{2}=\frac{1-\tan\left(\theta \right)}{1+\tan\left(\theta \right)},\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}$$
- step2: Simplify:
  $$\frac{\left(1-\tan\left(\theta \right)\right)^{2}}{\tan^{2}\left(\theta \right)}=\frac{1-\tan\left(\theta \right)}{1+\tan\left(\theta \right)}$$
- step3: Cross multiply:
  $$\left(1-\tan\left(\theta \right)\right)^{2}\left(1+\tan\left(\theta \right)\right)=\tan^{2}\left(\theta \right)\left(1-\tan\left(\theta \right)\right)$$
- step4: Move the expression to the left side:
  $$\left(1-\tan\left(\theta \right)\right)^{2}\left(1+\tan\left(\theta \right)\right)-\tan^{2}\left(\theta \right)\left(1-\tan\left(\theta \right)\right)=0$$
- step5: Solve using substitution:
  $$\left(1+\tan\left(\theta \right)\right)\left(1-\tan\left(\theta \right)\right)^{2}-\tan^{2}\left(\theta \right)\left(1-\tan\left(\theta \right)\right)=0$$
- step6: Rearrange the terms:
  $$1-\tan\left(\theta \right)-2\tan^{2}\left(\theta \right)+2\tan^{3}\left(\theta \right)=0$$
- step7: Factor the expression:
  $$\left(1-\tan\left(\theta \right)\right)\left(1-2\tan^{2}\left(\theta \right)\right)=0$$
- step8: Separate into possible cases:
  $$\begin{align}&1-\tan\left(\theta \right)=0\&1-2\tan^{2}\left(\theta \right)=0\end{align}$$
- step9: Solve the equation:
  $$\begin{align}&\tan\left(\theta \right)=1\&\tan\left(\theta \right)=\frac{\sqrt{2}}{2}\&\tan\left(\theta \right)=-\frac{\sqrt{2}}{2}\end{align}$$
- step10: Calculate:
  $$\begin{align}&\theta =\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\&\tan\left(\theta \right)=\frac{\sqrt{2}}{2}\&\tan\left(\theta \right)=-\frac{\sqrt{2}}{2}\end{align}$$
- step11: Calculate:
  $$\begin{align}&\theta =\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\&\theta =\arctan\left(\frac{\sqrt{2}}{2}\right)+k\pi ,k \in \mathbb{Z}\&\tan\left(\theta \right)=-\frac{\sqrt{2}}{2}\end{align}$$
- step12: Calculate:
  $$\begin{align}&\theta =\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\&\theta =\arctan\left(\frac{\sqrt{2}}{2}\right)+k\pi ,k \in \mathbb{Z}\&\theta =\arctan\left(-\frac{\sqrt{2}}{2}\right)+k\pi ,k \in \mathbb{Z}\end{align}$$
- step13: Find the union:
  $$\theta =\left\{ \begin{array}{l}\arctan\left(-\frac{\sqrt{2}}{2}\right)+k\pi \\arctan\left(\frac{\sqrt{2}}{2}\right)+k\pi \\frac{\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}$$
- step14: Check if the solution is in the defined range:
  $$\theta =\left\{ \begin{array}{l}\arctan\left(-\frac{\sqrt{2}}{2}\right)+k\pi \\arctan\left(\frac{\sqrt{2}}{2}\right)+k\pi \\frac{\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z},\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}$$
- step15: Find the intersection:
  $$\theta =\left\{ \begin{array}{l}\arctan\left(-\frac{\sqrt{2}}{2}\right)+k\pi \\arctan\left(\frac{\sqrt{2}}{2}\right)+k\pi \\frac{\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}$$
Let's solve the equation step by step.

The given equation is:

$$
\left(\frac{1 - \tan(\theta)}{\cos(\theta)}\right)^{2} = \frac{1 - \sin(\theta)}{1 + \sin(\theta)}
$$

### Step 1: Simplify the Left Side

The left side can be rewritten using the identity $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$:

$$
\left(\frac{1 - \frac{\sin(\theta)}{\cos(\theta)}}{\cos(\theta)}\right)^{2} = \left(\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta)}\right)^{2}
$$

This simplifies to:

$$
\left(\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta)}\right)^{2} = \left(1 - \frac{\sin(\theta)}{\cos(\theta)}\right)^{2} = \left(1 - \tan(\theta)\right)^{2}
$$

### Step 2: Simplify the Right Side

The right side can be simplified using the identity for the difference of squares:

$$
\frac{1 - \sin(\theta)}{1 + \sin(\theta)} = \frac{(1 - \sin(\theta))(1 - \sin(\theta))}{(1 + \sin(\theta))(1 - \sin(\theta))} = \frac{(1 - \sin(\theta))^2}{\cos^2(\theta)}
$$

### Step 3: Set the Two Sides Equal

Now we have:

$$
\left(1 - \tan(\theta)\right)^{2} = \frac{(1 - \sin(\theta))^2}{\cos^2(\theta)}
$$

### Step 4: Solve for $$\theta$$

From the previous calculations, we can derive the solutions for $$\theta$$:

$$
\theta = \left\{ 
\begin{array}{l}
\arctan\left(-\frac{\sqrt{2}}{2}\right) + k\pi \
\arctan\left(\frac{\sqrt{2}}{2}\right) + k\pi \
\frac{\pi}{4} + k\pi 
\end{array}
\right., k \in \mathbb{Z}
$$

### Final Answer

The solutions for $$\theta$$ are:

$$
\theta = \left\{ 
\begin{array}{l}
\arctan\left(-\frac{\sqrt{2}}{2}\right) + k\pi \
\arctan\left(\frac{\sqrt{2}}{2}\right) + k\pi \
\frac{\pi}{4} + k\pi 
\end{array}
\right., k \in \mathbb{Z}
$$
$$\theta = \left\{ \begin{array}{l} \arctan\left(-\frac{\sqrt{2}}{2}\right) + k\pi \ \arctan\left(\frac{\sqrt{2}}{2}\right) + k\pi \ \frac{\pi}{4} + k\pi \end{array} \right., k \in \mathbb{Z}$$

\( (\frac{1-\tan (\theta )}{\cos (\theta )})^{2}=\frac{1-\sin (\theta )}{1+\sin (\theta )} \)

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