Answer :
Let's solve the given equation:
[tex]\[ 2 \tan^{-1}(x) + \tan^{-1}(x + 1) = \frac{\pi}{2} \][/tex]
To start, let's denote [tex]\(\theta = \tan^{-1}(x)\)[/tex]. Therefore, [tex]\(x = \tan(\theta)\)[/tex]. This substitution will help us manipulate the equation more easily.
### Step 1: Rewriting the equation
Our equation becomes:
[tex]\[ 2\theta + \tan^{-1}(\tan(\theta) + 1) = \frac{\pi}{2} \][/tex]
### Step 2: Use the identity for arctan addition
Recall the addition formula for arctangents:
[tex]\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \][/tex]
However, here we cannot directly use that formula, but we'll break it down into simpler steps. Firstly, let's rewrite [tex]\( \tan^{-1}(\tan(\theta) + 1) \)[/tex]:
Let [tex]\(y = \tan^{-1}(x + 1)\)[/tex]. Then, given that [tex]\(\theta = \tan^{-1}(x)\)[/tex], and substituting [tex]\(y\)[/tex],
we have:
[tex]\[ y = \tan^{-1}(\tan(\theta) + 1) \][/tex]
### Step 3: Solve [tex]\(2\theta + y = \frac{\pi}{2}\)[/tex]
We know that [tex]\( y = \frac{\pi}{2} - 2\theta \)[/tex] because substituting [tex]\( y \)[/tex] into our original equation should simplify it:
[tex]\[ 2\theta + y = \frac{\pi}{2} \,\Rightarrow\, y = \frac{\pi}{2} - 2\theta \][/tex]
We now aim to find the relationship between [tex]\(x\)[/tex] and [tex]\(\theta\)[/tex] that satisfies this new equation:
[tex]\[ \tan^{-1}(x + 1) = \frac{\pi}{2} - 2\tan^{-1}(x) \][/tex]
### Step 4: Apply tangent on both sides
[tex]\[ x + 1 = \tan\left( \frac{\pi}{2} - 2\tan^{-1}(x)\right) \][/tex]
Using the identity [tex]\(\tan\left( \frac{\pi}{2}-A \right) = \cot(A)\)[/tex]:
[tex]\[ x + 1 = \cot\left( 2\tan^{-1}(x) \right) \][/tex]
### Step 5: Use double-angle identity for cotangent and arctangent:
Recall:
[tex]\[ \cot(2A) = \frac{1-\tan^2(A)}{2\tan(A)}: \text{where } A = \tan^{-1}(x) \][/tex]
So,
[tex]\[ x + 1 = \frac{1 - \tan^2(\tan^{-1}(x))}{2\tan(\tan^{-1}(x))} \][/tex]
[tex]\[ x + 1 = \frac{1 - x^2}{2x} \][/tex]
### Step 6: Solve the resulting equation
Re-write and arrange terms:
[tex]\[ x + 1 = \frac{1 - x^2}{2x} \][/tex]
Multiply both sides by [tex]\(2x\)[/tex]:
[tex]\[ 2x(x + 1) = 1 - x^2 \][/tex]
[tex]\[ 2x^2 + 2x = 1 - x^2 \][/tex]
[tex]\[ 3x^2 + 2x - 1 = 0 \][/tex]
### Step 7: Solve the quadratic equation
Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 12}}{6} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{16}}{6} \][/tex]
[tex]\[ x = \frac{-2 \pm 4}{6} \][/tex]
So,
[tex]\[ x = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ x = \frac{-6}{6} = -1 \][/tex]
### Conclusions
The solutions to the original equation are:
[tex]\[ x = \frac{1}{3} \, \text{and} \, x = -1 \][/tex]
Hence, the solutions are [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( x = -1 \)[/tex].
[tex]\[ 2 \tan^{-1}(x) + \tan^{-1}(x + 1) = \frac{\pi}{2} \][/tex]
To start, let's denote [tex]\(\theta = \tan^{-1}(x)\)[/tex]. Therefore, [tex]\(x = \tan(\theta)\)[/tex]. This substitution will help us manipulate the equation more easily.
### Step 1: Rewriting the equation
Our equation becomes:
[tex]\[ 2\theta + \tan^{-1}(\tan(\theta) + 1) = \frac{\pi}{2} \][/tex]
### Step 2: Use the identity for arctan addition
Recall the addition formula for arctangents:
[tex]\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \][/tex]
However, here we cannot directly use that formula, but we'll break it down into simpler steps. Firstly, let's rewrite [tex]\( \tan^{-1}(\tan(\theta) + 1) \)[/tex]:
Let [tex]\(y = \tan^{-1}(x + 1)\)[/tex]. Then, given that [tex]\(\theta = \tan^{-1}(x)\)[/tex], and substituting [tex]\(y\)[/tex],
we have:
[tex]\[ y = \tan^{-1}(\tan(\theta) + 1) \][/tex]
### Step 3: Solve [tex]\(2\theta + y = \frac{\pi}{2}\)[/tex]
We know that [tex]\( y = \frac{\pi}{2} - 2\theta \)[/tex] because substituting [tex]\( y \)[/tex] into our original equation should simplify it:
[tex]\[ 2\theta + y = \frac{\pi}{2} \,\Rightarrow\, y = \frac{\pi}{2} - 2\theta \][/tex]
We now aim to find the relationship between [tex]\(x\)[/tex] and [tex]\(\theta\)[/tex] that satisfies this new equation:
[tex]\[ \tan^{-1}(x + 1) = \frac{\pi}{2} - 2\tan^{-1}(x) \][/tex]
### Step 4: Apply tangent on both sides
[tex]\[ x + 1 = \tan\left( \frac{\pi}{2} - 2\tan^{-1}(x)\right) \][/tex]
Using the identity [tex]\(\tan\left( \frac{\pi}{2}-A \right) = \cot(A)\)[/tex]:
[tex]\[ x + 1 = \cot\left( 2\tan^{-1}(x) \right) \][/tex]
### Step 5: Use double-angle identity for cotangent and arctangent:
Recall:
[tex]\[ \cot(2A) = \frac{1-\tan^2(A)}{2\tan(A)}: \text{where } A = \tan^{-1}(x) \][/tex]
So,
[tex]\[ x + 1 = \frac{1 - \tan^2(\tan^{-1}(x))}{2\tan(\tan^{-1}(x))} \][/tex]
[tex]\[ x + 1 = \frac{1 - x^2}{2x} \][/tex]
### Step 6: Solve the resulting equation
Re-write and arrange terms:
[tex]\[ x + 1 = \frac{1 - x^2}{2x} \][/tex]
Multiply both sides by [tex]\(2x\)[/tex]:
[tex]\[ 2x(x + 1) = 1 - x^2 \][/tex]
[tex]\[ 2x^2 + 2x = 1 - x^2 \][/tex]
[tex]\[ 3x^2 + 2x - 1 = 0 \][/tex]
### Step 7: Solve the quadratic equation
Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 12}}{6} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{16}}{6} \][/tex]
[tex]\[ x = \frac{-2 \pm 4}{6} \][/tex]
So,
[tex]\[ x = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ x = \frac{-6}{6} = -1 \][/tex]
### Conclusions
The solutions to the original equation are:
[tex]\[ x = \frac{1}{3} \, \text{and} \, x = -1 \][/tex]
Hence, the solutions are [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( x = -1 \)[/tex].
