Q) For acute angles A and B, if sec (2 A – B) = √2 and cosec (A + B) = 2 then find the values of A and B.

(Q 22 A – 30/5/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Given that sec (2 A – B) = √2

∵ we know that, sec 45 ⁰ = √2

∴ 2 A – B = 45 ⁰ …………. (i)

and given that cosec ( A + B) = 2

∵ we know that, cosec 30 ⁰ = 2

∴ A + B = 30 ⁰

Step 2: Let’s solve the equations (i) and (ii) for values of A and B:

By adding both the equations, we get

(2 A – B) + ( A + B) = 45 ⁰ + 30 ⁰

3 A = 75 ⁰

∴ A = = 25 ⁰

Step 3: by substituting value of A in equation (ii), we get:

25 + B = 30 ⁰

∴ B = 30 ⁰ – 25 ⁰ = 5 ⁰

Therefore, values of A and B are 25 ⁰ and 5 ⁰ respectively.

(Caution Tip: Since angles have to be acute, hence values of angles A & B should be < 90 ⁰)

Please press “Heart” if you liked the solution.