Problem 57 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The positive integral solution of the equation \(\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^{2}}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)\) is (A) \(x=1, y=2\) (B) \(x=2, y=1\) (C) \(x=3, y=2\) (D) \(x=-2, y=-1\)

Step-by-Step Solution

Answer

The solution is (B) \(x=2, y=1\).

Step 1: Understand the Inverse Trigonometric Identities

Review the identities: \( \tan^{-1}(x) \) and \( \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \tan^{-1}(y) \). This is true because \( \cos(\tan^{-1}(y)) = \frac{1}{\sqrt{1+y^2}} \), which implies \( \frac{y}{\sqrt{1+y^2}} = \sin(\tan^{-1}(y)) \). Thus, \( \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \tan^{-1}(y) \).

Step 2: Simplify the Given Equation

Substitute the identities into the equation: \( \tan^{-1}(x) + \tan^{-1}(y) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \). The equation now involves the tangent addition formula.

Step 3: Use the Tangent Addition Formula

Apply the tangent addition formula: \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \). This transforms the equation into \( \tan^{-1}\left(\frac{x + y}{1 - xy}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \).

Step 4: Evaluate \(\sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \)

Observe that \( \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \) corresponds to an angle whose sine is \( \frac{3}{\sqrt{10}} \) and can be represented by \( \theta = \tan^{-1}(\frac{3}{1}) \), since \( \tan(\theta) = \frac{3}{\sqrt{10-9}} = 3 \).

Step 5: Set Equivalent Tangents

Equate the tangents from steps 3 and 4: \( \frac{x + y}{1 - xy} = 3 \). Solve for \( x \) and \( y \).

Step 6: Solve for Integer Solutions

Re-arrange and consider positive integer values: \( x + y = 3(1 - xy) \). Expanding gives \( x + y = 3 - 3xy \). Options show possible values for \( x \) and \( y \).

Step 7: Test Given Options

Test each option: (A) \(x=1, y=2\) or (B) \(x=2, y=1\). Substitute in \( \frac{x + y}{1 - xy} = 3 \). Both give valid results but check (A): \( \frac{1+2}{1-1\times2} = \frac{3}{-1} gtr 3 \). (B) works: \( \frac{2+1}{1-2\times1} = 3 \).

Step 8: Conclusion

The valid positive integer solution is only option (B).

Key Concepts

Tangent Addition Formula

The Tangent Addition Formula is an essential tool when dealing with inverse trigonometric functions and solving equations involving them. It states that:

• \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \) when \( ab < 1 \).
• It helps simplify complex expressions that otherwise might not easily be calculated.

For the exercise, applying this formula allowed us to transform an equation into a simpler form. By using this transformation, it became possible to match the left side of an equation to a known value on the right side. The formula is particularly useful in equations where \( \tan^{-1} \) appears as it translates the sum of two inverse tangents into the tangent of their addition. This simplification was key to solving the original trigonometric equation.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved variables. They serve as essential tools in simplifying expressions and solving complex trigonometric problems.
A few crucial identities to be aware of include:

• The identity: \( \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}} \), implies that values like \( \frac{y}{\sqrt{1+y^2}} \) simplify to \( \sin(\tan^{-1}(y)) \).
• Inverse relationships, such as converting between \( \cos^{-1} \) and \( \tan^{-1} \).

In the exercise, recognizing and substituting these identities was crucial. It helped transform the initial complex equation involving a mix of inverse trigonometric functions. By simplifying and reducing using such identities, it was possible to manage the flow towards a tangible solution. This method ensures an easier pathway when manipulating equations in trigonometric problem-solving.

Solution of Equations

Solving equations is about finding the values, or roots, that satisfy a given equation. In equations involving inverse trigonometric functions, recognizing equivalent forms and utilizing known identities are essential tactics.
Here’s how you can approach such equations:

• Substitute known identities to simplify complex terms. This often transforms convoluted expressions into basic forms that are easier to work with.
• Use algebraic manipulation to rearrange the equation into a solvable format, like shifting terms or factoring.
• Test possible solutions if specific integer or natural number results are required, as in many multiple-choice problems.

In our example, once the equation was simplified using identities and tangent transformations, necessary algebraic steps identified the integer solutions. Testing provided answer choices confirmed which pair satisfies the constraint set by the equation. This approach is not only systematic but also minimizes mistakes in trigonometric equation-solving.