Problem 18 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{2}\), then the value of \(q\) is (A) 1 (B) \(\frac{1}{\sqrt{2}}\) (C) \(\frac{1}{3}\) (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \( q \) is 1.

Step 1: Understanding the Equation

We are given the equation \( \cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} + \cos^{-1} \sqrt{1-q} = \frac{3\pi}{2} \). We need to find the value of \( q \).

Step 2: Using the Identity

Recall that \( \cos^{-1}(a) + \cos^{-1}(b) = \pi \) if \( a^2 + b^2 = 1 \). Apply this to \( \cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} = \pi \) because \( p + 1-p = 1 \).

Step 3: Substituting the Identity Result

Since \( \cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} = \pi \), substitute this result into the original equation to get \( \pi + \cos^{-1} \sqrt{1-q} = \frac{3\pi}{2} \).

Step 4: Simplifying the Equation

Subtract \( \pi \) from both sides of the equation: \( \cos^{-1} \sqrt{1-q} = \frac{\pi}{2} \).

Step 5: Solving for \( \sqrt{1-q} \)

If \( \cos^{-1} x = \frac{\pi}{2} \), then \( x = 0 \). So, \( \sqrt{1-q} = 0 \).

Step 6: Finding and Verifying \( q \)

Solving \( \sqrt{1-q} = 0 \) gives \( 1-q = 0 \). Thus, \( q = 1 \). This matches option (A).

Key Concepts

Cosine Function

The cosine function is a fundamental part of trigonometry. In our case, we are dealing with its inverse function, denoted as \( \cos^{-1} \). The inverse cosine function, \( \cos^{-1}(x) \), is used to find the angle whose cosine is \( x \). This function is defined for values of \( x \) ranging from \(-1\) to \(1\), producing an angle range of \([0, \pi]\). Understanding this function's range and output is critical when working through trigonometric equations.
Learning how these functions switch between angles and ratios is crucial for solving inverse trigonometric problems. In this problem, \( \cos^{-1} \sqrt{p} \) yields an angle, which, combined with others, results in a specific sum. Recognizing that each \( \cos^{-1} \) effectively translates a ratio back into an angle guides us in manipulating or solving trigonometric identities and equations.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold for all values of the variable where both sides of the equation are defined. One integral identity relevant here is \( \cos^{-1}(a) + \cos^{-1}(b) = \pi \) if \( a^2 + b^2 = 1 \). This identity helps simplify complex trigonometric expressions by reducing them to more manageable terms.
In this exercise, you can apply the identity to \( \cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} = \pi \). It simplifies the original equation, aligning these two terms with a known sum and, consequently, enabling a more straightforward path to finding the unknown variable \( q \). Such identities are powerful tools in breaking down and solving trigonometric problems.

Equation Solving

Equation solving in trigonometry often involves transforming the given equation into a simpler form using identities or inverse functions. Once simplified, the equation becomes easier to manipulate and solve for unknown variables. In this instance, simplifying the given equation using a trigonometric identity leads to:

• \( \pi + \cos^{-1} \sqrt{1-q} = \frac{3\pi}{2} \)

Subtracting \( \pi \) from both sides provides:

• \( \cos^{-1} \sqrt{1-q} = \frac{\pi}{2} \)

This implies that the \( \cos \) of the angle \( \frac{\pi}{2} \) is zero, so \( \sqrt{1-q} \) must also be zero. Knowing \( \cos^{-1}(x) = \frac{\pi}{2} \) results in \( x = 0 \), we find \( \sqrt{1-q} = 0 \), leading to \( 1-q = 0 \) and thus \( q = 1 \).
Such systematic solving lies at the heart of any trigonometric challenge: apply identities, simplify, solve, and verify the solution.