For the equation `2x=tan(2tan^(-1)a)+2 t a n(tan^(-1)a+tan^(-1)a^3)` , which of the following is/are invalid? a.`"a"^2"x"+2"a"="x"` b. `"a"^2+2"a x"+1=0` c.`"a"!=0` d. `a!=-1,\ 1`

Options

A. `a^(2)x+2a=x`
B. `a^(2)+2ax+1=0`
C. `a ne 0`
D. `a ne 1, -1 `

Answer

To solve the equation \( 2x = \tan(2 \tan^{-1} a) + 2 \tan(\tan^{-1} a + \tan^{-1} a^3) \) and determine which of the given options are invalid, we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ 2x = \tan(2 \tan^{-1} a) + 2 \tan(\tan^{-1} a + \tan^{-1} a^3) \]

### Step 2: Use the double angle formula for tangent
We know that:
\[ \tan(2A) = \frac{2\tan A}{1 - \tan^2 A} \]
Let \( A = \tan^{-1} a \). Then:
\[ \tan(2 \tan^{-1} a) = \frac{2a}{1 - a^2} \]

### Step 3: Use the addition formula for tangent
For \( \tan(A + B) \):
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( B = \tan^{-1} a^3 \). Then:
\[ \tan(\tan^{-1} a + \tan^{-1} a^3) = \frac{a + a^3}{1 - a \cdot a^3} = \frac{a + a^3}{1 - a^4} \]

### Step 4: Substitute back into the equation
Now substituting back, we have:
\[ 2x = \frac{2a}{1 - a^2} + 2 \cdot \frac{a + a^3}{1 - a^4} \]

### Step 5: Simplify the equation
Combining the terms:
\[ 2x = \frac{2a}{1 - a^2} + \frac{2(a + a^3)}{1 - a^4} \]

### Step 6: Find a common denominator
The common denominator for the two fractions is \((1 - a^2)(1 - a^4)\). Thus, we rewrite:
\[ 2x = \frac{2a(1 - a^4) + 2(a + a^3)(1 - a^2)}{(1 - a^2)(1 - a^4)} \]

### Step 7: Expand and simplify the numerator
Expanding the numerator:
\[ 2a(1 - a^4) + 2(a + a^3)(1 - a^2) = 2a - 2a^5 + 2a + 2a^3 - 2a^3 - 2a^5 = 4a - 4a^5 \]

Thus, we have:
\[ 2x = \frac{4a(1 - a^4)}{(1 - a^2)(1 - a^4)} \]

### Step 8: Simplify further
Dividing both sides by 2 gives:
\[ x = \frac{2a(1 - a^4)}{(1 - a^2)(1 - a^4)} \]

### Step 9: Cross-multiply
Cross-multiplying gives:
\[ x(1 - a^2)(1 - a^4) = 2a \]

### Step 10: Rearranging the equation
Rearranging leads to:
\[ x - x a^2 - x a^4 = 2a \]
This can be rearranged to:
\[ x a^2 + 2a - x = 0 \]

### Step 11: Analyze the options
Now we analyze the options provided:
1. \( a^2 x + 2a = x \) - This is valid.
2. \( a^2 + 2ax + 1 = 0 \) - This is invalid.
3. \( a \neq 0 \) - This is invalid since \( a = 0 \) is a solution.
4. \( a \neq -1, 1 \) - This is valid since \( a = 1 \) and \( a = -1 \) lead to undefined expressions in the original equation.

### Conclusion
Thus, the invalid options are:
- Option 2: \( a^2 + 2ax + 1 = 0 \)
- Option 3: \( a \neq 0 \)