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Question

If sec x – cos x = 4, then what will be the value of \({{(1 \ + \ cos^2 x)} \over cos x}\)?

This question was previously asked in
SSC CGL 2022 Tier-II (Paper 2 JSO) Previous Year Paper (04-Mar-2023)
The correct answer is \(2{ \sqrt{5}} \)

Solving Trigonometric Equations: Finding \(\frac{1 + \cos^2 x}{\cos x}\)

The problem asks us to find the value of the expression \(\frac{1 + \cos^2 x}{\cos x}\) given the equation \(\sec x - \cos x = 4\).

Breaking Down the Given Equation

We are given the equation:

\(\sec x - \cos x = 4\)

Recall the definition of the secant function: \(\sec x = \frac{1}{\cos x}\). Substitute this into the equation:

\(\frac{1}{\cos x} - \cos x = 4\)

This gives us a direct relationship between \(\frac{1}{\cos x}\) and \(\cos x\).

Analyzing the Expression to Find

The expression we need to evaluate is \(\frac{1 + \cos^2 x}{\cos x}\). We can split this expression into two terms:

\(\frac{1 + \cos^2 x}{\cos x} = \frac{1}{\cos x} + \frac{\cos^2 x}{\cos x}\)

Simplifying the second term, we get:

\(\frac{1}{\cos x} + \cos x\)

So, the problem reduces to finding the value of \(\frac{1}{\cos x} + \cos x\).

Connecting the Given Equation to the Expression

We know the value of \(\frac{1}{\cos x} - \cos x\), which is 4. We need to find the value of \(\frac{1}{\cos x} + \cos x\).

Let \(a = \frac{1}{\cos x}\) and \(b = \cos x\). We are given \(a - b = 4\) and we need to find \(a + b\).

We can use the algebraic identity: \((a+b)^2 = (a-b)^2 + 4ab\).

Substitute \(a = \frac{1}{\cos x}\) and \(b = \cos x\) into the identity:

\(\left(\frac{1}{\cos x} + \cos x\right)^2 = \left(\frac{1}{\cos x} - \cos x\right)^2 + 4 \left(\frac{1}{\cos x}\right) (\cos x)\)

We know that \(\frac{1}{\cos x} - \cos x = 4\). Also, \(4 \left(\frac{1}{\cos x}\right) (\cos x) = 4 \times 1 = 4\).

Substitute these values into the equation:

\(\left(\frac{1}{\cos x} + \cos x\right)^2 = (4)^2 + 4\)

\(\left(\frac{1}{\cos x} + \cos x\right)^2 = 16 + 4\)

\(\left(\frac{1}{\cos x} + \cos x\right)^2 = 20\)

Now, take the square root of both sides to find the value of \(\frac{1}{\cos x} + \cos x\):

\(\frac{1}{\cos x} + \cos x = \pm \sqrt{20}\)

Simplify the square root:

\(\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}\)

So, \(\frac{1}{\cos x} + \cos x = \pm 2\sqrt{5}\).

Since the expression we need to find is \(\frac{1}{\cos x} + \cos x\), its value is either \(2\sqrt{5}\) or \(-2\sqrt{5}\). Looking at the options provided, \(2\sqrt{5}\) is one of the options.

Step-by-Step Calculation

  • Start with the given equation: \(\sec x - \cos x = 4\).
  • Replace \(\sec x\) with \(\frac{1}{\cos x}\): \(\frac{1}{\cos x} - \cos x = 4\).
  • Identify the expression to evaluate: \(\frac{1 + \cos^2 x}{\cos x}\).
  • Rewrite the expression: \(\frac{1}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{1}{\cos x} + \cos x\).
  • Let \(A = \frac{1}{\cos x}\) and \(B = \cos x\). We have \(A - B = 4\) and need to find \(A + B\).
  • Use the identity \((A+B)^2 = (A-B)^2 + 4AB\).
  • Substitute the values: \(\left(\frac{1}{\cos x} + \cos x\right)^2 = \left(\frac{1}{\cos x} - \cos x\right)^2 + 4 \left(\frac{1}{\cos x}\right) (\cos x)\).
  • \(\left(\frac{1}{\cos x} + \cos x\right)^2 = (4)^2 + 4(1)\).
  • \(\left(\frac{1}{\cos x} + \cos x\right)^2 = 16 + 4 = 20\).
  • Take the square root: \(\frac{1}{\cos x} + \cos x = \pm \sqrt{20}\).
  • Simplify the square root: \(\sqrt{20} = 2\sqrt{5}\).
  • Thus, \(\frac{1 + \cos^2 x}{\cos x} = \frac{1}{\cos x} + \cos x = \pm 2\sqrt{5}\).

Comparing with the given options, the value is \(2\sqrt{5}\).

Revision Table: Key Trigonometry Concepts

Concept Definition/Identity Usage in Problem
Secant Function \(\sec x = \frac{1}{\cos x}\) Used to rewrite the given equation.
Algebraic Identity \((a+b)^2 = (a-b)^2 + 4ab\) Crucial for finding \(a+b\) from \(a-b\) and \(ab\).
Expression Manipulation Splitting fractions: \(\frac{X+Y}{Z} = \frac{X}{Z} + \frac{Y}{Z}\) Used to rewrite \(\frac{1 + \cos^2 x}{\cos x}\) as \(\frac{1}{\cos x} + \cos x\).
Square Root Simplification \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) Used to simplify \(\sqrt{20}\) to \(2\sqrt{5}\).

Additional Information: Reciprocal Identities and Algebraic Tricks

Trigonometric identities are fundamental in solving trigonometry problems. The reciprocal identities, like \(\sec x = \frac{1}{\cos x}\), \(\csc x = \frac{1}{\sin x}\), and \(\cot x = \frac{1}{\tan x}\), help in converting expressions into simpler forms involving sine and cosine.

In this problem, a key step was recognizing that if you have the difference of two quantities (\(a-b\)) and their product (\(ab\)), you can find their sum (\(a+b\)) using the identity \((a+b)^2 = (a-b)^2 + 4ab\). Here, the quantities were \(\frac{1}{\cos x}\) and \(\cos x\). Their product is \(\frac{1}{\cos x} \times \cos x = 1\). This simplification of the product is what made the algebraic identity particularly useful in this trigonometric context.

Always look for opportunities to simplify trigonometric expressions using identities and to apply standard algebraic techniques when dealing with trigonometric equations.

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