If \(2 \tan \alpha+\cot \beta=\tan \beta\) then \(\tan (\beta-\alpha)=\) (a) tana (b) cota (c) \(\tan \beta\) (d) \(\cot \beta\)

Understanding the relationship between tangent and cotangent functions is crucial for solving trigonometry problems. These functions are reciprocals of each other, which means \( \cot \theta = \frac{1}{\tan \theta} \) and conversely \( \tan \theta = \frac{1}{\cot \theta} \).

This reciprocal relationship allows us to interchange the functions when convenient, as seen in the original exercise where \( \cot \beta \) was rewritten in terms of \( \tan \beta \) to simplify the equation. Remembering these identities is key for solving more complex equations, especially when they involve multiple trigonometric functions. It's one of the trigonometry building blocks that students must master early on, especially in preparations for exams like JEE.

Trigonometric equations involve angles and sides of triangles or functions like sine, cosine, tangent, and their reciprocals. They can range from simple to complex, involving multiple steps and requiring a good understanding of trigonometric identities and formulas.

When working with these equations, it's vital to express all terms in a single trigonometric function when possible. This process was demonstrated in the textbook solution, where \( 2 \tan \alpha +\cot \beta \) ultimately was expressed only in terms of \( \tan \) and \(- \tan \) functions. Solving trigonometric equations is a common task in the JEE Maths syllabus, where a clear, systematic approach to manipulation and transformation of equations is essential. Furthermore, being adept at rearranging and simplifying these equations will greatly improve problem-solving skills and confidence.

The tan addition formula is instrumental in solving trigonometry problems involving the addition or subtraction of two angles. It is given by \( \tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \cdot \tan \beta} \) where the signs in the numerator and denominator are opposite. This formula is a powerful tool for simplifying expressions and solving equations.

In the given problem, this formula was used to decompose \( \tan(\beta - \alpha) \) into a more manageable form, allowing to replace the product of \( \tan \alpha \) and \( \tan \beta \) with the expression we found earlier. This demonstrates the usefulness of the tan addition formula in breaking down complex trigonometric expressions into simpler, solvable components.

JEE Trigonometry encompasses a broad array of topics and problem types designed to challenge a student's understanding and application of trigonometric concepts. The syllabus includes but is not limited to trigonometric identities, equations, functions, and their properties. It is imperative for students to have a strong grasp of these topics, as they are integral to solving higher-level math problems and are frequently tested on the JEE.

For JEE aspirants, mastering trigonometry requires learning various identities, solving complex equations, and applying them to different contexts. Trigonometry problems on the JEE often demand a combination of analytical skills and algebraic manipulation, as showcased in the exercise where algebraic rearrangements and applying the tan addition formula led to the correct solution. For effective preparation, practicing a wide variety of problems and understanding their underlying principles is essential.