If sec(t) = x⁄y, then cot(t) is equal to:

Correct — D. sec t = x⁄y means cos t = y⁄x, so draw the right triangle it describes: adjacent = y, hypotenuse = x.

By Pythagoras the third side is opposite = √(x² − y²).

cot t = adjacent ⁄ opposite
= y⁄√(x² − y²) → option (d)

The identity route lands in the same place: tan²t = sec²t − 1 = x²⁄y² − 1 = (x² − y²)⁄y², so tan t = √(x² − y²)⁄y and cot t is its reciprocal.

Why the others are wrong

• (a)
  The plus sign puts x and y on the two legs and √(x² + y²) on the hypotenuse. But sec t = x⁄y makes x the hypotenuse, so the missing side comes from x² − y².
• (b)
  Same wrong triangle as option (a), and the numerator is x — the hypotenuse. cot t is adjacent over opposite, and the hypotenuse is neither of those.
• (c)
  The triangle is right but the numerator is not. x is the hypotenuse, so x⁄√(x² − y²) is hypotenuse over opposite, which is cosec t rather than cot t.

sec t = hypotenuse ⁄ adjacent, so sec t = x⁄y hands you two of the three sides at once: adjacent y, hypotenuse x.

Pythagoras supplies the third, opposite = √(x² − y²), and every remaining ratio is then a matter of picking two sides.

The sign under the root is the whole question. Because sec names the hypotenuse, the missing side carries a difference of squares. Had the stem given tan t = x⁄y, x and y would both be legs and the missing side would be √(x² + y²) — which is what options (a) and (b) are built from.

Key facts

• sec t = hypotenuse ⁄ adjacent, so sec t = x⁄y puts y on the adjacent side and x on the hypotenuse.
• 1 + tan²t = sec²t, which gives tan t = √(x² − y²)⁄y here.
• cot t = adjacent ⁄ opposite = y⁄√(x² − y²), the reciprocal of that tan value.

Common traps

• Putting x and y on the two legs, which turns the third side into √(x² + y²)
• Answering cosec t, x over the root, instead of cot t
• Inverting tan and cot at the last line

No directly related past PYQ was found.