Question

Statement 1 \(\tan ^{-1} 2+\tan ^{-1} \frac{1}{2}=\frac{\pi}{2}\) and Statement 2 \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\) where \(\mathrm{x}>0, \mathrm{y}>0\) and \(\mathrm{xy}<1\)

Short answer

#tag_title# Short Answer By identifying x = 2 and y = 1/2 and applying Statement 2, we confirmed all conditions were satisfied and simplified the expression, proving that the sum of the inverse tangent of 2 and the inverse tangent of 1/2 equals π/2.

Step-by-step solution

Identify x and y

In this case, we have \(\mathrm{x}=2\) and \(\mathrm{y}=\frac{1}{2}\). Now we will use Statement 2 to get the sum of the inverse tangents and check if the conditions are satisfied.

Apply Statement 2 and Check Conditions

Using Statement 2, we have $$ \tan ^{-1}(2)+\tan ^{-1}\left(\frac{1}{2}\right)=\tan^{-1}\left(\frac{2+\frac{1}{2}}{1-2\cdot\frac{1}{2}}\right). $$ Now, let's check the conditions: \(\mathrm{x}>0\), \(\mathrm{y}>0\), and \(\mathrm{xy}<1\): 1. \(2>0\): True 2. \(\frac{1}{2}>0\): True 3. \(2\cdot\frac{1}{2}<1\): True All conditions are satisfied, so we can proceed to simplify the expression.

Simplify the Expression

Now, simplifying the expression on the right-hand side: $$ \tan^{-1}\left(\frac{2+\frac{1}{2}}{1-2\cdot\frac{1}{2}}\right) = \tan^{-1}\left(\frac{\frac{5}{2}}{1-1}\right) = \tan^{-1}\left(\frac{\frac{5}{2}}{0}\right). $$ Since the tangent function approaches infinity as its argument approaches π/2, this expression is equivalent to: $$ \tan^{-1}\left(\infty\right) = \frac{\pi}{2}. $$ Therefore, $$ \tan^{-1}(2) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}, $$ which proves Statement 1.

Key concepts

Tan Inverse Addition Formula

The tan inverse addition formula is a useful tool for combining two arctangent functions into a single expression. This formula allows you to simplify an expression such as \(\tan^{-1} x + \tan^{-1} y\) into \(\tan^{-1}\frac{x+y}{1-xy}\) under certain conditions. The given conditions—\(x>0\), \(y>0\), and \(xy<1\)—ensure the resulting expression is valid as they prevent the denominator from becoming zero or negative, which would complicate matters since the tangent function isn't defined for a division by zero or a negative under the square root.

Using this formula, the angle obtained corresponds to the angle whose tangent is \(\frac{x+y}{1-xy}\). It's a powerful tool when dealing with inverse trigonometric functions and simplifying trigonometric expressions. However, students must remember that this addition formula applies only to certain ranges of x and y, emphasizing the need to check these conditions before applying the formula to ensure a correct solution.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arctangent (\(\tan^{-1}\)), arcsine (\(\sin^{-1}\)), and arccosine (\(\cos^{-1}\)), are the inverse operations of their respective trigonometric functions. They are used to determine angles from the known value of the trigonometric function. The inverse trigonometric functions are widely used in various branches of mathematics and physics to solve problems involving angles and lengths in right-angled triangles.

Understanding these functions is crucial for simplifying trigonometric expressions and solving complex problems in higher-level mathematics. When working with inverse trigonometric identities, one should keep in mind the principal value range for each function, as this determines the domain for which the inverses are defined. For example, for the arctangent function, the principal value range is \( (-\pi/2, \pi/2) \) excluding the ends.

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions is a core skill in trigonometry, which forms an essential part of IIT-JEE Mathematics preparation and higher mathematics courses. It involves reducing complex trigonometric expressions into simpler forms using identities, formulas, and algebraic manipulation to make them easier to interpret or solve. Techniques such as factoring, finding common denominators, and using fundamental trigonometric identities are crucial.

Effective simplification often requires a deep understanding of trigonometric identities, including Pythagorean identities, angle sum and difference identities, double angle identities, and, importantly, inverse trigonometric properties. Simplification can not only make equations easier to solve but also aids in the visualization of trigonometric functions and the understanding of their properties.

IIT-JEE Mathematics

IIT-JEE Mathematics is considered one of the most challenging sections of the Joint Entrance Examination, which is a standardized test used for admission to engineering colleges in India, including the prestigious Indian Institutes of Technology (IITs). The mathematics section of IIT-JEE tests a broad range of topics from algebra, geometry, trigonometry to calculus, and differential equations.

Concepts like the tan inverse addition formula and simplifying trigonometric expressions are components of the trigonometry segment covered in the IIT-JEE syllabus. Students aiming to excel in IIT-JEE Mathematics should focus on conceptual clarity, problem-solving skills, and application of various mathematical theorems and formulas. Moreover, practice with a wide range of problems, adherence to the prescribed syllabus, and time management are critical for success in this competitive examination.