If $ 2 \tan A = 3 $, then the value of $ (\sec A + \tan A - 1)... | Mathematics MCQ | Examoogle

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If $2 \tan A = 3$ , then the value of $(\sec A + \tan A - 1)(\sec A - \tan A + 1)$ is:

$\frac{2}{3}$

$\frac{1}{3}$

Correct Answer

📐 MA • Math Algebraic SimplificationMathematicsTrigonometry

Option A

Quick Summary:

Recognize that $(\sec A + \tan A - 1)(\sec A - \tan A + 1)$ is in the form $(X - Y)(X + Y)$ if rearranged as $(\sec A + (\tan A - 1))(\sec A - (\tan A - 1))$ , which simplifies to $\sec^2 A - (\tan A - 1)^2$ .

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📐MAMath SolutionAlgebraic Simplification

📋 Given

The equation given is 2 tan A = 3. We need to evaluate the expression (sec A + tan A - 1)(sec A - tan A + 1).

🔢 Formula Used

1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A1+tan2A=sec2A

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Students often try to solve for $\tan A = 1.5$ and then calculate $\sec A = \sqrt{1 + 2.25} = 1.6$ , which is prone to decimal errors; using the identity directly is safer.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Simplify the expression

Rewrite the expression as $(\sec A + (\tan A - 1))(\sec A - (\tan A - 1))$ . This matches the difference of squares identity $(x+y)(x-y) = x^2 - y^2$ , where $x = \sec A$ and $y = \tan A - 1$ .

sec⁡2A−(tan⁡A−1)2\sec^2 A - (\tan A - 1)^2sec2A−(tanA−1)2

Apply trigonometric identity

Expand the squared term and substitute $\sec^2 A = 1 + \tan^2 A$ .

(1+tan⁡2A)−(tan⁡2A−2tan⁡A+1)=1+tan⁡2A−tan⁡2A+2tan⁡A−1(1 + \tan^2 A) - (\tan^2 A - 2\tan A + 1) = 1 + \tan^2 A - \tan^2 A + 2\tan A - 1(1+tan2A)−(tan2A−2tanA+1)=1+tan2A−tan2A+2tanA−1

Calculate final result

Simplify the equation, the result is $2 \tan A$ . Since we are given $2 \tan A = 3$ , the value is 3.

2tan⁡A=32 \tan A = 32tanA=3

✅

A is correct because the expression simplifies algebraically to $2 \tan A$ , which is given as 3.

Core Concepts Used

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Trigonometric Identities Difference of Squares Algebraic Simplification

💡 EXAM TIP

This type of algebraic manipulation in trigonometry is very common in competitive exams to avoid calculating irrational square roots of secant or cosecant.

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