The given expression is equivalent to: √((sec θ + 1) ⁄ (sec θ − 1)) + √((sec θ − 1) ⁄ (sec θ + 1))

Correct — A. Put both surds over the single denominator √((sec θ + 1)(sec θ − 1)):

= [(sec θ + 1) + (sec θ − 1)] ⁄ √(sec²θ − 1)
= 2 sec θ ⁄ √(sec²θ − 1)

sec²θ − 1 = tan²θ, so the denominator is tan θ:
= 2 sec θ ⁄ tan θ

Now write both ratios in sine and cosine:
= 2 × (1 ⁄ cos θ) × (cos θ ⁄ sin θ)
= 2 ⁄ sin θ = 2 cosec θ → option (a).

Why the others are wrong

• (b)
  2 tan θ sec θ multiplies by tan θ where the working divides by it. tan θ arrives as the denominator √(sec²θ − 1), so it sits under the 2 sec θ.
• (c)
  2 sin θ confuses cosec with sin. The simplification ends at 2 ⁄ sin θ, and 2 ⁄ sin θ is 2 cosec θ, the reciprocal ratio, not sin θ doubled.
• (d)
  2 tan θ keeps the tan θ from √(sec²θ − 1) and throws away the 2 sec θ above it. The numerator does not cancel — it is divided by tan θ.

Two square roots that are reciprocals of each other always add the same way: √(A⁄B) + √(B⁄A) = (A + B) ⁄ √(AB). One move clears both radicals, which is faster than rationalising the two terms separately.

Here A = sec θ + 1 and B = sec θ − 1, so A + B = 2 sec θ and AB = sec²θ − 1 = tan²θ. The difference of squares in the denominator is the whole reason the pair was chosen.

The three Pythagorean identities behind this family are sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.

Key facts

• sec²θ − 1 = tan²θ, the rearranged form of 1 + tan²θ = sec²θ.
• √(A⁄B) + √(B⁄A) = (A + B) ⁄ √(AB).
• sec θ ÷ tan θ = 1 ⁄ sin θ = cosec θ.

Common traps

• Rationalising each surd on its own and losing a sign in the second term.
• Writing sec²θ − 1 as cot²θ by picking the wrong Pythagorean identity.
• Stopping at 2 sec θ ⁄ tan θ and choosing whichever option contains tan θ.

No directly related past PYQ was found.