Q) Let 2A + B and A + 2B be acute angles such that sin (2A + B) = √3 / 2 and tan (A + 2B) = 1. Find the value of cot (4A – 7B).

PYQ: 30 (b) – CBSE 2025 – Code 30 – Series 5 – Set 1

Ans: Let’s start solving given conditions one by one:

Step 1: Since sin (2 A + B) = √3 / 2

we know that sin 60⁰ = √3 / 2

∴ sin (2 A + B) = sin 60⁰

∴ 2 A + B = 60⁰ …………… (i)

Step 2: Since tan (A + 2 B) = 1

we know that tan 45⁰ = 1

∴ tan (A + 2 B) = tan 45⁰

∴ A + 2 B = 45⁰ …………… (ii)

Step 3: Let’s find values of A and B by solving equations (i) and (ii):

We multiply equation (i) by 2, we get:

2 A + B = 60⁰ …………… (i)

∴ (2 A + B) 2 = 60⁰ x 2

∴ 4 A + 2 B = 120⁰ ……… (iii)

Next, we subtract equation (ii) from equation (iii), we get:

(4 A + 2 B) – (A + 2 B) = 120⁰ – 45⁰

∴ 4 A + 2 B – A – 2 B = 75⁰

∴ 3 A = 75⁰

∴ A = 25⁰

Next, by substituting value of A in equation (i), we get:

2 A + B = 60⁰ ………….. (i)

∴ 2 (25⁰) + B = 60⁰

∴ 50⁰ + B = 60⁰

∴ B = 60⁰ – 50⁰

∴ B = 10⁰

Step 4: Next, we find the value of (4A – 7B):

4 A – 7 B = 4 (25⁰) – 7 (10⁰) = 100⁰ – 70⁰ = 30⁰

∴ Value of cot (4 A – 7 B) = cot 30⁰ = √3

Therefore, the value of cot (4 A – 7 B) is √3.

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