If \(\tan ^{-1} 2 x+\tan ^{-1} 3 x=(\pi / 4)\) then its solution is (a) \(\\{1,(1 / 6)\\}\) (b) \(\\{\pm(1 / 6)\\}\) (c) \(\\{-1,(1 / 6)\\}\) (d) \(\\{1 / 6\\}\)

The addition formula for tangent is a fundamental concept in trigonometry used to find the tangent of the sum of two angles. If you have two angles, denoted as A and B, then the formula is:
\[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}\].
This equation is especially useful when dealing with inverse trigonometric functions, such as when you encounter a problem that requires you to add two angles and the tangents of these angles are known.
In the case of the exercise, we are working with the angles whose tangents are 2x and 3x, respectively. Knowing that the sum of their arctangents equals \(\frac{\pi}{4}\), which is a special angle whose tangent value is 1, allows us to set up an equation. Once we use the addition formula and simplify, we can proceed to solve for x. This step is crucial as it directs us to understand the relationship between two seemingly complex expressions.

Inverse trigonometric functions allow us to find angles given a trigonometric ratio. Among their numerous properties, one that is often relied upon is \(\tan(\tan^{-1}(x)) = x\).

This property is particularly helpful when solving equations involving inverse trigonometric functions. It simplifies expressions and allows us to work with algebraic values rather than trigonometric ones. In the context of our exercise, we used this property to simplify \(\tan(\tan^{-1}(2x))\) to 2x, and \(\tan(\tan^{-1}(3x))\) to 3x, making the equation much more approachable. Understanding how inverse trigonometric functions behave and being able to quickly convert between angles and ratios is crucial for working through problems involving these functions.

Why Is This Useful?

Mastering these properties grants students the ability to translate complex trigonometric expressions into simple algebraic forms, thereby making the process of finding solutions more straightforward.

A quadratic equation is an equation of the second degree, which means it includes at least one term that is squared. The general form of a quadratic equation is \(ax^2 + bx + c = 0\).
To solve such equations, one can factorize, complete the square, or use the quadratic formula. In our step-by-step solution, we obtained a quadratic equation \(6x^2 + 5x - 1 = 0\) after applying trigonometric and algebraic manipulations.

The solutions of a quadratic equation can be real or complex, and they represent the values of x that will satisfy the original equation. For the given exercise, solving the quadratic equation helps find the value of x that meets the condition set by the combination of inverse trigonometric functions previously established.

Factoring vs. Quadratic Formula

While factoring is a straightforward method, it does not always yield an easily discernible solution. Conversely, the quadratic formula is a fail-safe approach that works for all quadratic equations, greatly simplifying the problem-solving process when the equation does not factor nicely.