SSC CGL 2024 Tier-I · 25 September 2024, Shift 2 · Quantitative Aptitude · Question 5 of 99

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.

Find the value of tan 72° − tan 27° − tan 72° tan 27°.

(a)−1
(b)0
(c)1
(d)−2

Answer

Why

Correct — C. Read the angles before the expression: 72° − 27° = 45°.

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B)

With A = 72° and B = 27°, the left side is tan 45° = 1.

So (tan 72° − tan 27°) ⁄ (1 + tan 72° tan 27°) = 1
Cross-multiplying: tan 72° − tan 27° = 1 + tan 72° tan 27°

Move tan 72° tan 27° across:
tan 72° − tan 27° − tan 72° tan 27° = 1 → option (c)

Why the others are wrong

(a)
−1 — −1 is tan(−45°), which you reach by running the identity as tan(27° − 72°). The expression is anchored to tan 72° − tan 27°, so the difference is +45° and the constant is +1.
(b)
0 — 0 would require tan 72° = tan 27° + tan 72° tan 27°. Numerically the three terms are 3.0777 − 0.5095 − 1.5682 = 1, not 0.
(d)
−2 — −2 needs both a sign flip and a factor of 2, and the identity supplies neither. After cross-multiplying, the right-hand side is the bare constant 1.

Concept

The whole question is the compound-angle identity for tangent:

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B)

When a stem hands you two angles whose difference is 45°, cross-multiply the identity rather than evaluating either tangent.

tan 45° = 1 turns it into tan A − tan B = 1 + tan A tan B, and the printed expression is that same line with tan A tan B carried to the left.

The mirror version, for angles that sum to 45°, is tan A + tan B + tan A tan B = 1.

Neither tan 72° nor tan 27° is a standard value and no calculator is allowed. That absence is the signal to look at the difference of the angles instead of the angles themselves.

Key facts

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B).
If A − B = 45°, then tan A − tan B − tan A tan B = 1.
If A + B = 45°, then tan A + tan B + tan A tan B = 1.
tan 45° = 1, and here 72° − 27° = 45°.

Study next

The companion identity for two angles summing to 45 degreesCompound-angle formulae for sine and cosine used the same wayComplementary pairs such as tan θ × tan(90° − θ) = 1

Common traps

Hunting for a decimal value of tan 72° instead of noticing that 72° − 27° = 45°
Cross-multiplying and then dropping the 1, which is the term that survives, and answering 0
Reaching for the sum formula, which needs a plus sign between the tangents in the numerator

SSC builds this item backwards from an angle pair that differs by, or sums to, 45 degrees. Check the pair before you read the expression: the identity collapses to a constant and no individual tangent value is ever needed.

Related PYQs

No directly related past PYQ was found.

Find the value of tan 72° − tan 27° − tan 72° tan 27°.

(a)−1
(b)0
(c)1
(d)−2

Answer

Why

Correct — C. Read the angles before the expression: 72° − 27° = 45°.

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B)

With A = 72° and B = 27°, the left side is tan 45° = 1.

So (tan 72° − tan 27°) ⁄ (1 + tan 72° tan 27°) = 1
Cross-multiplying: tan 72° − tan 27° = 1 + tan 72° tan 27°

Move tan 72° tan 27° across:
tan 72° − tan 27° − tan 72° tan 27° = 1 → option (c)

Why the others are wrong

(a)
−1 — −1 is tan(−45°), which you reach by running the identity as tan(27° − 72°). The expression is anchored to tan 72° − tan 27°, so the difference is +45° and the constant is +1.
(b)
0 — 0 would require tan 72° = tan 27° + tan 72° tan 27°. Numerically the three terms are 3.0777 − 0.5095 − 1.5682 = 1, not 0.
(d)
−2 — −2 needs both a sign flip and a factor of 2, and the identity supplies neither. After cross-multiplying, the right-hand side is the bare constant 1.

Concept

The whole question is the compound-angle identity for tangent:

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B)

When a stem hands you two angles whose difference is 45°, cross-multiply the identity rather than evaluating either tangent.

tan 45° = 1 turns it into tan A − tan B = 1 + tan A tan B, and the printed expression is that same line with tan A tan B carried to the left.

The mirror version, for angles that sum to 45°, is tan A + tan B + tan A tan B = 1.

Neither tan 72° nor tan 27° is a standard value and no calculator is allowed. That absence is the signal to look at the difference of the angles instead of the angles themselves.

Key facts

tan(A − B) = (tan A − tan B) ⁄ (1 + tan A tan B).
If A − B = 45°, then tan A − tan B − tan A tan B = 1.
If A + B = 45°, then tan A + tan B + tan A tan B = 1.
tan 45° = 1, and here 72° − 27° = 45°.

Study next

The companion identity for two angles summing to 45 degreesCompound-angle formulae for sine and cosine used the same wayComplementary pairs such as tan θ × tan(90° − θ) = 1

Common traps

Hunting for a decimal value of tan 72° instead of noticing that 72° − 27° = 45°
Cross-multiplying and then dropping the 1, which is the term that survives, and answering 0
Reaching for the sum formula, which needs a plus sign between the tangents in the numerator

Related PYQs

No directly related past PYQ was found.