SSC CGL 2024 Tier-I · 12 September 2024, Shift 3 · Quantitative Aptitude · Question 6 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 6 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 tan 4θ. tan6θ = 1, where 6θ is an acute angle, then find the value of Cot5θ.

Answer

Why

Correct — D. The stem and all four options are printed as images: the question reads 'If tan 4θ · tan 6θ = 1, where 6θ is an acute angle, then find the value of cot 5θ', and the options show √3, −√3, −1 and 1 for option (a) through option (d).

Rule: tan A · tan B = 1 means A and B are complementary, because tan(90° − A) = cot A.

So 4θ + 6θ = 90°
10θ = 90° → θ = 9°

Check the rider: 6θ = 54°, which is acute, as the question demands.

cot 5θ = cot 45° = 1 → option (d).

Why the others are wrong

(a)
√3 is cot 30°, which needs 5θ = 30° and θ = 6°. Then tan 24° · tan 36° comes to about 0.32, nowhere near 1.
(b)
−√3 is cot 150°, needing θ = 30°. That makes 6θ = 180°, so tan 6θ = 0 and the product is 0 — and 180° is not acute either.
(c)
−1 is cot 135°, giving θ = 27°. The product tan 108° · tan 162° does equal 1, so the equation alone cannot rule this out — but 6θ = 162° is obtuse, and the rider requires it acute.

Concept

tan and cot are cofunctions: tan(90° − A) = cot A. So tan A · tan B = 1 is the same statement as tan B = cot A, which holds when A + B = 90°.

That is the working rule, but it is not the complete solution set. Because tangent has period 180°, the equation actually holds whenever A + B = 90° + 180°k. With A = 4θ and B = 6θ, k = 0 gives θ = 9° and k = 1 gives θ = 27°.

The restriction that 6θ is acute is what selects θ = 9° and makes cot 5θ single-valued.

This row carries no text at all — the question and the four options are images, and the option images read √3, −√3, −1 and 1 in order. Both negative options correspond to real roots of the equation that the acute-angle rider excludes, so the rider is doing genuine work rather than tidying up.

Key facts

tan(90° − A) = cot A, so tan A · tan B = 1 whenever A + B = 90°.
The general solution is A + B = 90° + 180°k, which is why a range restriction is needed for uniqueness.
cot 45° = 1, cot 30° = √3, cot 135° = −1 and cot 150° = −√3.
With θ = 9°, the three angles in play are 4θ = 36°, 5θ = 45° and 6θ = 54°.

Study next

Complementary-angle identities: sin(90° − A) = cos A, tan(90° − A) = cot ASolving trigonometric equations under a stated angle restrictionStandard values of cot at 30°, 45°, 60°, 135° and 150°

Common traps

Ignoring the acute-angle rider, which leaves θ = 27° as a second root and −1 as a second answer.
Treating tan A · tan B = 1 as A = B, which is true only at 45°.
Reaching θ = 9° and then reporting tan 5θ or θ itself instead of cot 5θ.

SSC writes this family as a product set to 1 with multiple angles inside, so the item collapses the moment you read it as complementary angles rather than as an equation to expand. A plain standard-value item appears at Quant Q.17, which asks for the exact value of sin 150°.

Related PYQs

No directly related past PYQ was found.