SSC CGL 2024 Tier-I · 23 September 2024, Shift 3 · Quantitative Aptitude · Question 20 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.

For what value of t is the value of sin²(t), half of the value of tan (t)?

(a)30°
(b)45°
(c)60°
(d)22.5°

Answer

Why

Correct — B. The stem is printed as an image: find t for which sin²t is half of tan t.

Write tan t as sin t⁄cos t:
sin²t = sin t⁄(2 cos t)

Divide both sides by sin t (t is not 0° among the options):
sin t = 1⁄(2 cos t)
2 sin t cos t = 1

The left side is the double-angle form:
sin 2t = 1
2t = 90°
t = 45° → option (b)

Check by substitution: sin²45° = 1⁄2, tan 45° = 1, and half of 1 is 1⁄2.

Why the others are wrong

(a)
30° — At 30°, sin²30° = 0.25 while half of tan 30° is 0.2887. Close, but not equal — and sin 2t = sin 60° = 0.866, not 1.
(c)
60° — At 60°, sin²60° = 0.75 while half of tan 60° is 0.866. The two sides move apart above 45°, because tan grows much faster than sin².
(d)
22.5° — 22.5° is 45° halved. It is what you write down if you reach sin 2t = 1 and then solve 2t = 45° instead of 2t = 90°. At 22.5° the two sides are 0.1464 and 0.2071.

Concept

A trig equation mixing sin²t with tan t is solved by putting everything over cos t and then looking for a double-angle shape.

The pivot here is sin 2t = 2 sin t cos t. Once the equation reduces to 2 sin t cos t = 1, it is really sin 2t = 1, and sin equals 1 only at 90° in the first revolution.

Dividing by sin t is legal only because sin t ≠ 0 for every option offered. Note what it costs: t = 0° satisfies the original equation (both sides are 0) and is lost the moment you divide.

Reading the stem backwards as "tan t is half of sin²t" gives 2 = sin t cos t, so sin 2t = 4, which is impossible. That dead end is a quick way to confirm you have the two sides the right way round.

Key facts

sin 2t = 2 sin t cos t, so sin t cos t = 1⁄2 is the same statement as sin 2t = 1.
sin θ = 1 first occurs at θ = 90°, so 2t = 90° and t = 45°.
At 45°: sin 45° = 1⁄√2, so sin²45° = 1⁄2, and tan 45° = 1.
t = 0° also satisfies sin²t = ½ tan t, since both sides vanish there.

Study next

The double-angle identities sin 2t, cos 2t and tan 2tSolving sin θ = 1, cos θ = 1 and tan θ = 1 over 0° to 360°The standard-angle table for 0°, 30°, 45°, 60° and 90°

Common traps

Solving 2t = 45° instead of 2t = 90°, which puts 22.5° straight into your answer.
Cancelling sin t without noticing that this discards the trivial solution t = 0°.
Squaring the whole equation to clear the fraction, which introduces roots that do not satisfy the original.

When the options are standard angles, substituting each one is a quick check on the algebra. An expression to be evaluated at θ = 45° is asked at Quant Q.16 of this paper.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 86 SSC CGL questions on it, every solved shift.