Question

$\frac{1+x}{1-x} \theta \leftarrow x$

Answer 1

$\tan\left(\frac{\pi}{4} + \theta\right)$

Answer 2

The provided input "$\frac{1+x}{1-x} \theta \leftarrow x$" is ambiguous due to the non-standard notation "$\theta \leftarrow x$". However, in the context of JEE/NEET exams, the expression $\frac{1+x}{1-x}$ frequently appears in problems involving trigonometric substitutions, particularly when dealing with inverse trigonometric functions or complex numbers.

The most common and relevant interpretation for this type of problem is to assume that the notation implies a relationship between $x$ and $\theta$, specifically a substitution that simplifies the expression. The standard substitution for the form $\frac{1+x}{1-x}$ is $x = \tan \theta$.

Solution:

Let's assume the implied task is to simplify the expression $\frac{1+x}{1-x}$ by substituting $x = \tan \theta$.

1. Substitute $x = \tan \theta$ into the expression: $\frac{1+x}{1-x} = \frac{1+\tan \theta}{1-\tan \theta}$

2. Recognize the trigonometric identity:

   The expression $\frac{1+\tan \theta}{1-\tan \theta}$ is a standard trigonometric identity derived from the tangent addition formula, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. We know that $\tan(\frac{\pi}{4}) = 1$. So, we can rewrite the expression as: $\frac{1+\tan \theta}{1-\tan \theta} = \frac{\tan(\frac{\pi}{4}) + \tan \theta}{1 - \tan(\frac{\pi}{4})\tan \theta}$

3. Apply the tangent addition formula: $\frac{\tan(\frac{\pi}{4}) + \tan \theta}{1 - \tan(\frac{\pi}{4})\tan \theta} = \tan\left(\frac{\pi}{4} + \theta\right)$

Therefore, under the substitution $x = \tan \theta$, the expression $\frac{1+x}{1-x}$ simplifies to $\tan\left(\frac{\pi}{4} + \theta\right)$.

Explanation of the solution:

The expression $\frac{1+x}{1-x}$ is a common form in trigonometry. By substituting $x = \tan \theta$, it transforms into $\frac{1+\tan \theta}{1-\tan \theta}$. This is a direct application of the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, with $A = \frac{\pi}{4}$ (since $\tan(\frac{\pi}{4})=1$) and $B = \theta$. Thus, the expression simplifies to $\tan\left(\frac{\pi}{4} + \theta\right)$.