If tanA = x – 1/4x, then what will be the value of secA – tanA?

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RRB NTPC Exam 19 Jan 2021 click here to view all questions

If tanA = x – 1/4x, then what will be the value of secA – tanA?

A$$2x$$

B$$x$$

C$$\frac{1}{2x}$$

D$$\frac{1}{2x}^2$$

Akhilesh ? Apr 18 '25 at 04:16

correct answer is: option c (1/2x)
Explanation:
Given: \(\tan A = x - \frac{1}{4x}\)

We need to find: \(\sec A - \tan A\)

**Steps:**

1. Express \(\tan A\) with a common denominator: \[ \tan A = \frac{4x^2 - 1}{4x} \]

2. Use the identity \(1 + \tan^2 A = \sec^2 A\): \[ \sec^2 A = 1 + \left(\frac{4x^2 - 1}{4x}\right)^2 = \frac{16x^4 + 8x^2 + 1}{16x^2} \]

3. Take the square root to find \(\sec A\): \[ \sec A = \frac{4x^2 + 1}{4x} \]

4. Subtract \(\tan A\) from \(\sec A\): \[ \sec A - \tan A = \frac{4x^2 + 1}{4x} - \frac{4x^2 - 1}{4x} = \frac{2}{4x} = \frac{1}{2x} \]

**Final Answer:**

\[ \boxed{\frac{1}{2x}} \]

All DiscussionsClick here to write answer

correct answer is: option c (1/2x) Explanation: Given: \(\tan A = x - \frac{1}{4x}\) We need to find: \(\sec A - \tan A\) **Steps:** 1. Express \(\tan A\) with a common denominator: \[ \tan A = \frac{4x^2 - 1}{4x} \] 2. Use the identity \(1 + \tan^2 A = \sec^2 A\): \[ \sec^2 A = 1 + \left(\frac{4x^2 - 1}{4x}\right)^2 = \frac{16x^4 + 8x^2 + 1}{16x^2} \] 3. Take the square root to find \(\sec A\): \[ \sec A = \frac{4x^2 + 1}{4x} \] 4. Subtract \(\tan A\) from \(\sec A\): \[ \sec A - \tan A = \frac{4x^2 + 1}{4x} - \frac{4x^2 - 1}{4x} = \frac{2}{4x} = \frac{1}{2x} \] **Final Answer:** \[ \boxed{\frac{1}{2x}} \] Akhilesh ? answered Apr 18 '25 at 04:16

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