If (sec α + tan α) ⁄ (sec α − tan α) = 7⁄4, then find the value of cosec α.

Correct — C. The stem is an image: (sec α + tan α) ⁄ (sec α − tan α) = 7⁄4, find cosec α.

Put both functions over cos α:
sec α + tan α = (1 + sin α) ⁄ cos α
sec α − tan α = (1 − sin α) ⁄ cos α
cos α cancels, so the ratio is (1 + sin α) ⁄ (1 − sin α) = 7⁄4

Cross-multiply: 4(1 + sin α) = 7(1 − sin α)
4 + 4 sin α = 7 − 7 sin α
11 sin α = 3, so sin α = 3⁄11

cosec α = 1 ⁄ sin α = 11⁄3 → option (c)

Why the others are wrong

• (a)
  Cosec α is 1⁄sin α, so it can never sit between −1 and 1. Option (a), 1⁄3, would require sin α = 3, which is outside sine's range.
• (b)
  Same range failure. 1⁄11 demands sin α = 11 — it is the 11 of 11 sin α = 3 inverted before the 3 was divided through.
• (d)
  3⁄11 is sin α itself, the value the algebra reaches one line before the end. The stem asks for cosec α, so it still has to be turned upside down.

Two facts run this item, and one of them is much faster.

The identity sec²α − tan²α = 1 makes (sec α + tan α) and (sec α − tan α) reciprocals of each other, so multiplying the given ratio by that product gives (sec α + tan α)² = 7⁄4. That route works but drags surds along.

The exam route: write both functions over cos α. The denominators cancel and the ratio collapses to (1 + sin α) ⁄ (1 − sin α), a linear equation in sin α.

Two lines later sin α = 3⁄11, and the question only ever wanted its reciprocal.

Key facts

• sec²α − tan²α = 1, so sec α + tan α and sec α − tan α are reciprocals.
• sec α + tan α = (1 + sin α)⁄cos α and sec α − tan α = (1 − sin α)⁄cos α.
• Their ratio is (1 + sin α)⁄(1 − sin α), which turns 7⁄4 into a linear equation.
• Here sin α = 3⁄11, so cosec α = 11⁄3.

Common traps

• Stopping at sin α = 3⁄11 and answering that instead of its reciprocal.
• Cross-multiplying with the 4 and 7 swapped, which yields sin α = −3⁄11.
• Solving for sec α and tan α separately with surds when the sin-only route is two lines.

No directly related past PYQ was found.