Solve the following:

Correct — C. Add the two fractions and the identity 1 − cos²θ = sin²θ finishes the job.

1⁄(1 + cosθ) + 1⁄(1 − cosθ)
= [(1 − cosθ) + (1 + cosθ)] ⁄ [(1 + cosθ)(1 − cosθ)]
= 2 ⁄ (1 − cos²θ)

1 − cos²θ = sin²θ, so
= 2⁄sin²θ = 2cosec²θ → option (c)

The cosθ terms cancel in the numerator and the denominator is a difference of squares — that is the entire trick.

Why the others are wrong

• (a)
  2sin — sin ends up in the denominator, not the numerator: the sum is 2⁄sin²θ. Its reciprocal is cosec, so an answer written on sin has the fraction upside down.
• (b)
  2sec — sec is 1⁄cos, but no cosine survives. The cosθ terms cancel when the numerators are added, and (1 + cosθ)(1 − cosθ) leaves 1 − cos²θ, which is sin²θ.
• (d)
  2tan — tan = sin⁄cos keeps a cosine in play, and there is none left after the difference of squares. The simplified form carries sin alone.

Two moves settle almost every add-these-two-trig-fractions item.

Pair the conjugates: 1 + cosθ and 1 − cosθ multiply to the difference of squares 1 − cos²θ.

Convert with the Pythagorean identity: sin²θ + cos²θ = 1 gives 1 − cos²θ = sin²θ, and equally 1 − sin²θ = cos²θ.

After that it is only reciprocal vocabulary — 1⁄sinθ is cosecθ, 1⁄cosθ is secθ, 1⁄tanθ is cotθ. Doing the algebra correctly and then naming the wrong reciprocal is an easy mark to lose.

Key facts

• (1 + cosθ)(1 − cosθ) = 1 − cos²θ = sin²θ.
• 1⁄sinθ = cosecθ, so 2⁄sin²θ is 2cosec²θ.
• The same pairing done on sine gives 1⁄(1 + sinθ) + 1⁄(1 − sinθ) = 2sec²θ.

Common traps

• Adding the denominators as well as the numerators, which gives 2⁄2 = 1.
• Reaching 2⁄sin²θ and then writing 2sin²θ, flipping the reciprocal at the last step.
• Dropping the square and answering with a first-power cosec.

No directly related past PYQ was found.