Problem 16 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\tan (A-B)=1\) and \(\sec (A+B)=\frac{2}{\sqrt{3}}\), then find the smallest +ve values of \(A\) and \(B\) and their most general values.

Step-by-Step Solution

Answer

The smallest positive values are: \(A = \pi/12\) and \(B = \pi/3\). The most general values are: \(A = n\pi + \pi/12\) and \(B = 2n\pi + \pi/3\) for any integer \(n\).

Step 1: Solve for A-B using tan(A-B)=1

The equation \(\tan(A-B)=1\) implies that \(A-B\) is equal to the angle whose tangent is 1. This occurs at 45 degrees or \(\pi/4\) radians. So, \(A-B = n\pi + \pi/4\) where \(n\) is any integer.

Step 2: Solve for A+B using sec(A+B)=2/√3

The equation \(\sec(A+B)=2/\sqrt{3}\) implies that \(A+B\) is equal to the angle whose secant is \(2/\sqrt{3}\). Alternatively, this is the angle whose cosine is \(\sqrt{3}/2\), which occurs at 30 degrees or \(\pi/6\) radians. Thus, \(A+B = 2n\pi \pm \pi/6\) where \(2n\pi\) represents any full circle (since cosine is positive in both 1st and 4th quadrants, we use both \(\pm\)), and \(n\) is any integer.

Step 3: Solve for A and B

We can use the results from steps 1 and 2 to solve for A and B. This can be done by adding and subtracting our two equations to get: \(A = [(n\pi + \pi/4) + (2n\pi \pm \pi/6)]/2\) and \(B = [(2n\pi \pm \pi/6) - (n\pi + \pi/4)]/2\). By simplifying these equations, we can find out the smallest positive and the most general values of A and B. Remember to make sure that A and B fall within the range of 0 to 2\(\pi\) as they are angles in a triangle.

Key Concepts

Tangent Identity

The tangent identity is a fundamental concept in trigonometry that allows us to connect angles to their tangent values. When solving \(\tan(A-B) = 1\), we're essentially looking for an angle whose tangent value equals 1. In basic trigonometry, \(\tan(\theta) = 1\) at \(45^\circ\) or \(\frac{\pi}{4}\) radians. This is because the tangent function, which is the ratio of the opposite side to the adjacent side in a right triangle, reaches the value 1 when these sides are equal.

Because tangent is periodic with a period of \(\pi\), the general solution includes all angles that can be expressed as \(n\pi + \frac{\pi}{4}\), where \(n\) is an integer. This means every half-circle rotation from \(\frac{\pi}{4}\) (like \(\frac{5\pi}{4}, \frac{9\pi}{4}\), etc.) would satisfy \(\tan(x) = 1\).

Understanding the tangent identity is crucial for solving trigonometric equations involving differences of angles like \(A - B\), as in this case.

Secant and Cosine Relationship

Secant and cosine are closely related trigonometric functions. In fact, secant is the reciprocal of cosine. Therefore, solving \(\sec(A+B) = \frac{2}{\sqrt{3}}\) is equivalent to finding \(\cos(A+B)\) where the reciprocal is true, i.e., \(\cos(A+B) = \frac{\sqrt{3}}{2}\).

This cosine value corresponds to well-known angles: \(30^\circ\) (\(\frac{\pi}{6}\)) and \(330^\circ\) (\(\frac{11\pi}{6}\)) because the cosine function repeats every \(2\pi\) and is positive in both the first and fourth quadrants.

The general solution thus involves both angles adjusted by any full circle rotations of \(2n\pi\). This gives the equation \(A+B = 2n\pi \pm \frac{\pi}{6}\), where "\pm" accounts for the cyclic nature of the cosine function in two quadrants.

Recognizing this relationship between secant and cosine can help in solving equations where one function is expressed in terms of the other, like in this problem.

Angle Sums and Differences

The angle sum and difference identities are powerful tools in trigonometry that help simplify complex expressions into more manageable parts. In this exercise, we deal with the angles \(A - B\) and \(A + B\).

Using the results from the equations \(\tan(A-B) = 1\) which equals \(n\pi + \frac{\pi}{4}\), and \(\sec(A+B) = \frac{2}{\sqrt{3}}\), or in cosine form, \(A + B = 2n\pi \pm \frac{\pi}{6}\), we can explore these two basic trigonometric identities.

To solve for the angles \(A\) and \(B\), addition and subtraction can be employed effectively. By adding the equations, you isolate \(A\): \[2A = (n+2n)\pi + \frac{\pi}{4} \pm \frac{\pi}{6}\] leading to unique values for \(A\). Meanwhile, subtracting them isolates \(B\), allowing simplification: \[2B = (2n-n)\pi \pm \frac{\pi}{6} - \frac{\pi}{4}\].

Mastering these identities is key for solving not only this problem but a variety of other trigonometric problems where angle sums and differences appear, offering a strategy for untangling complicated trigonometric expressions.