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Simplify the given expression. (sin A + cosec A) 2 + (cos A − sec A) 2 + 3

$1+ \tan^2 A + \cot^2 A$

$\tan^2 A + \cot^2 A$

$3 + \tan^2 A + \cot^2 A$

$6 + \tan^2 A + \cot^2 A$

Correct Answer

📐 MA • Math Direct FormulaMathematicsPrevious Year Paper 2024

Option D

$6 + \tan^2 A + \cot^2 A$

Quick Summary:

Use the identity $\sin A \csc A = 1$ and $\cos A \sec A = 1$ . The expression simplifies to $1 + \csc^2 A + \sec^2 A - 2 + 3 = 2 + (1 + \cot^2 A) + (1 + \tan^2 A) = 4 + \cot^2 A + \tan^2 A$ . Wait, checking the simplification: $1 + \csc^2 A + \sec^2 A = 1 + (1 + \cot^2 A) + (1 + \tan^2 A) = 3 + \cot^2 A + \tan^2 A$ . Adding the remaining terms brings it to $6 + \tan^2 A + \cot^2 A$ .

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📐MAMath SolutionDirect Formula

📋 Given

Expression: (sin A + cosec $A)^2$ + (cos A - sec $A)^2$ + 3

🔢 Formula Used

(a+b)2+(c−d)2+3=sin⁡2A+csc⁡2A+2sin⁡Acsc⁡A+cos⁡2A+sec⁡2A−2cos⁡Asec⁡A+3(a+b)^2 + (c-d)^2 + 3 = \sin^2 A + \csc^2 A + 2\sin A \csc A + \cos^2 A + \sec^2 A - 2\cos A \sec A + 3(a+b)2+(c−d)2+3=sin2A+csc2A+2sinAcscA+cos2A+sec2A−2cosAsecA+3

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Students often forget that $\sin A \csc A = 1$ and end up expanding terms incorrectly or failing to convert to $\tan^2 A$ and $\cot^2 A$ .

🔢 Step-by-Step Solution

Expand the squares

Expanding the given expression $( \sin A + \csc A )^2 + ( \cos A - \sec A )^2 + 3$ using $(a+b)^2 = a^2+b^2+2ab$ and $(a-b)^2 = a^2+b^2-2ab$ .

sin⁡2A+csc⁡2A+2sin⁡Acsc⁡A+cos⁡2A+sec⁡2A−2cos⁡Asec⁡A+3\sin^2 A + \csc^2 A + 2\sin A \csc A + \cos^2 A + \sec^2 A - 2\cos A \sec A + 3sin2A+csc2A+2sinAcscA+cos2A+sec2A−2cosAsecA+3

Simplify trigonometric identities

Using $\sin A \csc A = 1$ , $\cos A \sec A = 1$ , and $\sin^2 A + \cos^2 A = 1$ .

(sin⁡2A+cos⁡2A)+csc⁡2A+sec⁡2A+2(1)−2(1)+3=1+csc⁡2A+sec⁡2A+3(\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2(1) - 2(1) + 3 = 1 + \csc^2 A + \sec^2 A + 3(sin2A+cos2A)+csc2A+sec2A+2(1)−2(1)+3=1+csc2A+sec2A+3

Convert to tan and cot

Using the identities $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$ .

1+(1+cot⁡2A)+(1+tan⁡2A)+3=6+tan⁡2A+cot⁡2A1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 3 = 6 + \tan^2 A + \cot^2 A1+(1+cot2A)+(1+tan2A)+3=6+tan2A+cot2A

✅

D is correct because simplifying the expression results in 6 + tan² A + cot² A.

Core Concepts Used

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Trigonometric Identities Algebraic Expansion Pythagorean Identities

💡 EXAM TIP

This type of simplification frequently appears in SSC and NDA exams; always look to group $\sin^2 A + \cos^2 A$ first as it immediately reduces complexity.

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