SSC CGL 2025 Tier-I · 14 September 2025, Shift 3 · Quantitative Aptitude · Question 25 of 95

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 13 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

If sin A + cos A = √2 sin A, then what is the value of tanA?

(a)√2
(b)1
(c)√2 + 1
(d)√2 − 1

Answer

Why

Correct — C.
Divide every term by cos A: tan A + 1 = √2 tan A
Collect the tan A terms: 1 = √2 tan A − tan A = (√2 − 1) tan A
Divide by (√2 − 1): tan A = 1 ⁄ (√2 − 1)

Rationalise with (√2 + 1): (√2 + 1) ⁄ ((√2 − 1)(√2 + 1))
Simplify the denominator: (√2)² − 1² = 2 − 1 = 1
So tan A = √2 + 1 → option (c)

Why the others are wrong

(a)
√2 — tan A = √2 fails tan A + 1 = √2 tan A: the left side is √2 + 1 ≈ 2.41, the right side is √2 × √2 = 2.
(b)
1 — tan A = 1 means A = 45°. Then sin A + cos A = √2, but √2 sin A = √2 × 1⁄√2 = 1, so the equation fails.
(d)
√2 − 1 — √2 − 1 is cot A, not tan A: cos A = (√2 − 1) sin A gives cos A ⁄ sin A = √2 − 1. Take the reciprocal and rationalise to reach √2 + 1.

Concept

When an equation holds only sin A and cos A, each to the first power, divide through by one of them to leave a single ratio. Dividing by cos A gives tan A directly, since sin A ⁄ cos A = tan A and cos A ⁄ cos A = 1.

The result 1 ⁄ (√2 − 1) then needs rationalising: multiply top and bottom by the conjugate √2 + 1, because (√2 − 1)(√2 + 1) = 2 − 1 = 1.

tan A = √2 + 1 ≈ 2.414 is tan 67.5°. A check: sin 67.5° ≈ 0.924 and cos 67.5° ≈ 0.383 add to 1.307, and √2 × 0.924 ≈ 1.307.

Key facts

Dividing an equation in sin A and cos A by cos A (cos A ≠ 0) turns it into one in tan A.
1 ⁄ (√2 − 1) = √2 + 1, because (√2 − 1)(√2 + 1) = 1.
tan 67.5° = √2 + 1 and tan 22.5° = √2 − 1.
cot A = 1 ⁄ tan A, so here cot A = √2 − 1.

Study next

sin A + cos A = √2 cos A, which gives tan A = √2 − 1Squaring sin A + cos A = k to find sin A cos A, using (sin A + cos A)² = 1 + 2 sin A cos ARationalising denominators of the form 1 ⁄ (√a − √b)

Common traps

Stopping at cos A ⁄ sin A = √2 − 1 and marking it as tan A.
Rationalising with (√2 − 1) instead of its conjugate (√2 + 1), which leaves a surd in the denominator.

The stem sets sin A + cos A equal to a multiple of one ratio and asks for tan A. The same divide-and-rationalise route settles 17 Sep 2024, 16:00, Quant Q.13, where sin θ + cos θ = √3 cos θ gives cot θ = (√3 + 1) ⁄ 2.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 131 SSC CGL questions on it, every solved shift.