The given expression simplifies to 2 csc A, so the correct option is (B) 2 cosec A.

Introduction

Simplifying trigonometric expressions like sin A / (1 + cos A) + (1 + cos A) / sin A is a common question in school exams and competitive tests. Mastering such problems strengthens understanding of trigonometric identities and helps in quickly identifying the correct option in multiple-choice questions.

Step-by-step Simplification

Given:
sin A / (1 + cos A) + (1 + cos A) / sin A

Take the LCM of the two fractions:

[sin² A + (1 + cos A)²] / [sin A (1 + cos A)]

Here, the numerator becomes sin² A + (1 + cos A)².

Expand (1 + cos A)²:

(1 + cos A)² = 1 + 2 cos A + cos² A

So the numerator is

sin² A + 1 + 2 cos A + cos² A

Use the Pythagorean identity sin² A + cos² A = 1:

sin² A + cos² A = 1 ⇒ sin² A + cos² A + 1 = 2

Thus the numerator becomes

2 + 2 cos A = 2 (1 + cos A)

Substitute back into the fraction:

[2 (1 + cos A)] / [sin A (1 + cos A)]

Cancel (1 + cos A) from numerator and denominator:

2 / sin A = 2 csc A

So, the simplified value is 2 csc A.

Explanation of Each Option

The options are:

• (A) 2 sec A
• (B) 2 cosec A
• (C) sec A
• (D) cosec A

Option (A) 2 sec A

sec A = 1 / cos A.
The simplified expression becomes 2 csc A = 2 / sin A, which depends on sin A, not cos A, so this option is incorrect.

Option (B) 2 cosec A

cosec A = 1 / sin A.
The final simplification gives 2 / sin A = 2 cosec A, so this option matches exactly and is correct.

Option (C) sec A

This is half of 2 sec A and still involves cos A rather than sin A, so it does not agree with the derived result 2 csc A.

Option (D) cosec A

This has the right trigonometric function but the wrong coefficient; the expression simplifies to 2 cosec A, not cosec A alone, so this is also incorrect.

Why the Identity is Useful

This identity,

sin A / (1 + cos A) + (1 + cos A) / sin A = 2 csc A,

is a standard trigonometric simplification used in many exam problems and proofs.

Recognizing patterns like sin² A + cos² A = 1 and using algebraic expansion helps solve similar expressions quickly and accurately.