$$ \frac { \cos x } { \operatorname { cosec } + \sin x } - \frac { \cos x } { \cos x \sin x } = \operatorname { tas } 2 x $$

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UpStudy AI · Accepted Answer

The equation given is:

$$
\frac{\cos x}{\csc x + \sin x} - \frac{\cos x}{\cos x \sin x} = \tan 2x
$$

Let's break this down step by step.

### Step 1: Simplify the Left Side

1. **Convert $$\csc x$$**:
   $$
   \csc x = \frac{1}{\sin x}
   $$
   Therefore, the left side becomes:
   $$
   \frac{\cos x}{\frac{1}{\sin x} + \sin x} - \frac{\cos x}{\cos x \sin x}
   $$

2. **Combine the terms in the denominator**:
   $$
   \frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x}
   $$
   Thus, the first term becomes:
   $$
   \frac{\cos x \sin x}{1 + \sin^2 x}
   $$

3. **Rewrite the second term**:
   $$
   \frac{\cos x}{\cos x \sin x} = \frac{1}{\sin x}
   $$

4. **Combine the two fractions**:
   $$
   \frac{\cos x \sin x}{1 + \sin^2 x} - \frac{1}{\sin x}
   $$

### Step 2: Find a Common Denominator

The common denominator for the two fractions is $$(1 + \sin^2 x) \sin x$$:
$$
\frac{\cos x \sin^2 x - (1 + \sin^2 x)}{(1 + \sin^2 x) \sin x}
$$

### Step 3: Set the Left Side Equal to the Right Side

Now we have:
$$
\frac{\cos x \sin^2 x - (1 + \sin^2 x)}{(1 + \sin^2 x) \sin x} = \tan 2x
$$

### Step 4: Express $$\tan 2x$$

Recall that:
$$
\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}
$$

### Step 5: Solve the Equation

At this point, the equation is quite complex. To solve it, we can either simplify further or use numerical methods. Given the complexity, I will attempt to solve it numerically for specific values of $$x$$ or use a graphical approach.

Let's try to find specific solutions for $$x$$ by evaluating the left and right sides for common angles.

### Step 6: Evaluate for Specific Values

Let's evaluate for $$x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$$.

1. **For $$x = 0$$**:
   - Left: $$\frac{0}{1} - \frac{0}{0} = 0$$
   - Right: $$\tan 0 = 0$$

2. **For $$x = \frac{\pi}{4}$$**:
   - Left: $$\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} - \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}} - 1 = 0$$
   - Right: $$\tan \frac{\pi}{2}$$ (undefined)

3. **For $$x = \frac{\pi}{2}$$**:
   - Left: Undefined
   - Right: Undefined

4. **For $$x = \frac{3\pi}{4}$$**:
   - Left: $$\frac{-\frac{\sqrt{2}}{2}}{0} - \frac{-\frac{\sqrt{2}}{2}}{0} = 0$$
   - Right: $$\tan \frac{3\pi}{2}$$ (undefined)

5. **For $$x = \pi$$**:
   - Left: $$0$$
   - Right: $$0$$

### Conclusion

The equation has solutions at $$x = 0$$ and $$x = \pi$$. Further analysis or numerical methods may be required for other potential solutions.
The equation has solutions at $$ x = 0 $$ and $$ x = \pi $$.

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