Problem 32 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of the equation \(\tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)\) is (A) \(x=\pm \frac{\sqrt{7}}{3}\) (B) \(x=\pm \frac{\sqrt{5}}{3}\) (C) \(x=\pm \frac{3 \sqrt{5}}{2}\) (D) none of these

Step-by-Step Solution

Answer

The solution is (B) \(x = \pm \frac{\sqrt{5}}{3}\).

Step 1: Understanding the Inverse Functions

The problem involves inverse trigonometric functions, specifically \( \cos^{-1}(x) \) and \( \cot^{-1}\left(\frac{1}{2}\right) \). It's important to understand that \( \cos^{-1}(x) \) represents an angle whose cosine is \( x \), and \( \cot^{-1}(\cdot) \) represents an angle whose cotangent is the given value.

Step 2: Expression of Arc Cosine as Angle

We interpret \( \theta = \cos^{-1}(x) \). This implies \( \cos(\theta) = x \). In a right triangle where the angle is \( \theta \), one side (adjacent) would be \( x \) and the hypotenuse would be 1.

Step 3: Convert Tangent Expression

The tangent of this angle \( \theta \) is \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). If the adjacent side is \( x \) and the hypotenuse is 1, the opposite side is \( \sqrt{1-x^2} \), so \( \tan(\theta) = \frac{\sqrt{1-x^2}}{x} \).

Step 4: Inverse Cotangent Evaluation

Next, evaluate \( \sin\left(\cot^{-1}\left(\frac{1}{2}\right)\right) \). Let \( \phi = \cot^{-1}\left(\frac{1}{2}\right) \), so \( \cot(\phi) = \frac{1}{2} \). In a right triangle, the cotangent is the adjacent over the opposite, so adjacent = 1, opposite = 2, and hypotenuse = \( \sqrt{1^2 + 2^2} = \sqrt{5} \).

Step 5: Compute Sine of Arc Cotangent

Thus, \( \sin(\phi) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \).

Step 6: Equating and Solving for x

Now, equate \( \tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x} \) to \( \sin(\cot^{-1}(\frac{1}{2})) = \frac{2\sqrt{5}}{5} \) and solve: \[ \frac{\sqrt{1-x^2}}{x} = \frac{2\sqrt{5}}{5} \].Cross-multiply to obtain: \( 5\sqrt{1-x^2} = 2x\sqrt{5} \).Squaring both sides yields \( 25(1-x^2) = 20x^2 \).

Step 7: Forming and Solving a Quadratic

Simplify the equation to find the solutions: \( 25 - 25x^2 = 20x^2 \) which simplifies to \( 45x^2 = 25 \). Solving gives \( x^2 = \frac{5}{9} \), so \( x = \pm \frac{\sqrt{5}}{3} \).

Key Concepts

Trigonometric Equations

Trigonometric equations involve finding the values of variables that satisfy certain trigonometric relationships. In our problem, the trigonometric equation we need to solve is \(\tan \left(\cos^{-1} x\right)=\sin \left(\cot^{-1}\frac{1}{2}\right)\). The strategy involves interpreting the equation in terms of angles and mainly requires understanding the properties of the inverse trigonometric functions involved.
Such equations often require transforming the problem using fundamental trigonometric identities or properties of inverse functions to create solvable algebraic equations.
In our equation, by separate evaluations of the inverse trigonometric functions involved and understanding their geometric meanings through reference angles, we can equate and solve for the unknown x. We first express each side in terms of trigonometric angles and equate their computations.

Inverse Function Properties

Inverse trigonometric functions, like \(\cos^{-1}\) and \(\cot^{-1}\), are used to find angles when the trigonometric values are known.

• For example, \(\cos^{-1}(x)\) is an angle \(\theta\) in the range \([0, \pi]\) whose cosine is \(x\).
• Similarly, \(\cot^{-1}\left(\frac{1}{2}\right)\) is an angle whose cotangent value is \(\frac{1}{2}\).

Understanding these angles in a right triangle allows us to use their geometric interpretations effectively.
For \(\cos^{-1}(x)\), we visualize a right triangle where the adjacent side is \(x\) and the hypotenuse is 1, enabling us to use Pythagorean theorem for the opposite side.
On the other hand, \(\cot^{-1}(\frac{1}{2})\) can be interpreted with a right triangle where the adjacent side is 1, the opposite is 2, and the hypotenuse is \(\sqrt{5}\). This provides a direct path to deduce that the sine of this angle is \(\frac{2\sqrt{5}}{5}\).

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions that hold true for all values of the involved variables. In our exercise, these identities help express complex trigonometric expressions in simpler forms.
When working with \(\tan(\theta)\) and \(\sin(\phi)\), leveraging Pythagorean identities is crucial.

• For a given angle \(\theta\), if we know \(\cos(\theta) = x\), then \(\sin(\theta) = \sqrt{1-x^2}\), which comes from the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).
• Similarly, for an angle \(\phi\) with known \(\cot(\phi)\), we deduce other side lengths in the triangle using \(1 + \cot^2(\phi) = \csc^2(\phi)\).

These identities allow us to replace one function with another, creating a simplified equation. In solving our main equation, we used these identities to transform each side to a form in which they could be equated and solved for \(x\). The algebraic manipulation post identities enforces the connection between different trigonometric aspects of the same element.