SSC CGL 2024 Tier-I · 24 September 2024, Shift 2 · Quantitative Aptitude · Question 19 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

If tanA + cotA = 2, then the value of tan²A + cot²A is:

(a)8
(b)4
(c)16
(d)2

Answer

Why

Correct — D. The question is printed as an image: if tanA + cotA = 2, find tan²A + cot²A.

Square the given sum:
(tanA + cotA)² = 2² = 4
tan²A + 2·tanA·cotA + cot²A = 4

The product is fixed — tanA · cotA = 1 — so the middle term is 2 × 1 = 2.

tan²A + cot²A = 4 − 2 = 2 → option (d)

Why the others are wrong

(a)
8 — The square of the given sum is 4, and every legitimate route starts there. 8 is twice that square, and nothing in the expansion multiplies by 2 — the middle term is subtracted, not doubled.
(b)
4 — 4 is (tanA + cotA)², not tan²A + cot²A. Squaring leaves a middle term 2·tanA·cotA = 2 sitting inside it, and that has to come off before you are looking at the sum of the two squares.
(c)
16 — 16 is 4², the given sum squared twice. One squaring is enough: tanA + cotA = 2 becomes 4, and a second squaring answers a question the paper never asked.

Concept

The identity being tested is (x + y)² = x² + y² + 2xy, run backwards: you are handed x + y and asked for x² + y².

What makes the trigonometric version quick is that the product is already known. tanA · cotA = 1 wherever both are defined, because cotA = 1⁄tanA.

So the middle term is 2 no matter what A is, and the whole question collapses to tan²A + cot²A = (tanA + cotA)² − 2. The same move works on sinA with cosecA, on cosA with secA, and on any x + 1⁄x.

There is a second route worth knowing. tanA + cotA = 2 says t + 1⁄t = 2, which forces t = 1 and so A = 45°. Then tan²A + cot²A = 1 + 1 = 2 — the same answer, and a fast check when you distrust the algebra.

Key facts

tanA · cotA = 1 wherever both are defined, because cotA = 1⁄tanA.
(x + y)² = x² + y² + 2xy, so x² + y² = (x + y)² − 2xy.
tan²A + cot²A = (tanA + cotA)² − 2.
t + 1⁄t = 2 has the single solution t = 1, so tanA + cotA = 2 pins A at 45°.

Study next

Sum-and-product identities: reaching x² + y² and x³ + y³ from x + yThe minimum of x + 1⁄x for positive x, which is why tanA + cotA = 2 pins A at 45° in the first quadrantExact values of the six ratios at 45°

Common traps

Stopping at 4, the square of the given sum, without removing the middle term.
Carrying tanA·cotA through as an unknown instead of replacing it with 1.
Squaring a second time and arriving at 16.

SSC builds these from one given expression and asks for a second, reached by squaring and then using a reciprocal or Pythagorean identity. The same squaring step is wanted at Quant Q.20 of this paper, where (sinA + cosecA)² + (cosA + secA)² has to be expanded.

Related PYQs

No directly related past PYQ was found.