If 0 < A < B < π, sin A - sin B = (1)/(2) and cos A - cos B = (3)/(2) , then the value of A + B is equal to

Correct answer: D. (4π)/(3) (sin A - sin B) ^ 2 + (cos A - cos B) ^ 2 = (1)/(2) + (3)/(2) = 2 ⇒ 2 - 2(sin Asin B + cos Acos B) = 2 ⇒ (B - A) = 0 ⇒ B - A = (π)/(2) ⇒ B = A + (π)/(2) sin A - sin B = 1 2 ⇒ sin A - cos A = 1 2 ⇒ sin A. 1 2 - cos A. 1 2 = (1)/(2) ⇒ (A - (π)/(4)) = sin(π)/(6) ⇒ A = (π)/(6) + (π)/(4) = (5π)/(12) A + B = A + A + (π)/(2) = (5π)/(6) + (π)/(2) = (4π)/(3)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If 0 < A < B < π, sin A - sin B = (1)/(2) and cos A - cos B =...

0 < \text{A} < \text{B} < \pi, sin A - sin B = \dfrac{1}{2}

and

cos A - cos B = \sqrt{\dfrac{3}{2}}

, then the value of

A + B

is equal to

Options

A $\dfrac{2\pi}{3}$
B $\dfrac{5\pi}{6}$
C $\pi$
D $\dfrac{4\pi}{3}$

Correct answer

D. $\dfrac{4\pi}{3}$

Step-by-step solution

{\left(sin A - sin B\right)}^{2} + {\left(cos A - cos B\right)}^{2} = \dfrac{1}{2} + \dfrac{3}{2} = 2

⇒ 2 - 2\left(sin Asin B + cos Acos B\right) = 2

⇒ \cos \left(B - A\right) = 0 ⇒ B - A = \dfrac{\pi}{2} ⇒ B = A + \dfrac{\pi}{2}

sin A - sin B = \dfrac{1}{\sqrt{2}} ⇒ sin A - cos A = \dfrac{1}{\sqrt{2}}

⇒ sin A. \dfrac{1}{\sqrt{2}} - cos A. \dfrac{1}{\sqrt{2}} = \dfrac{1}{2}

⇒ \sin \left(A - \dfrac{\pi}{4}\right) = sin\dfrac{\pi}{6}

⇒ A = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{5\pi}{12}

A + B = A + A + \dfrac{\pi}{2} = \dfrac{5\pi}{6} + \dfrac{\pi}{2} = \dfrac{4\pi}{3}

Practice Trigonometric Equations on Quantrex Academy →

More from Trigonometric Equations

Match each entry in List-I to the correct entry in List-II and choose the correct option.… JEE Advanced 2026 Passage: Consider the curve C_1 given by y = e^ -x for x ∈ [0, 10π] , and the curve C_2… JEE Advanced 2026 Let S = \ θ ∈ (-2π, 2π) : θ + 1 = 3 θ\ . Then ∑_ θ ∈ S θ is equal to: JEE Main 2026 If S = \ θ ∈ [-π, π] : θ (5θ)/(2) = 7θ (7θ)/(2)\ , then n(S) is equal to _______. JEE Main 2026 Let S = \ x ∈ [-π, π] : x ( x + x) = a, a ∈ ℤ\ . Then n(S) is equal to : JEE Main 2026 The number of elements in the set \ x ∈[0,180^ ∘ ]: (x+100^ ∘ )= (x+50^ ∘ ) x (x-50^ ∘ )\… JEE Main 2026 Number of solutions of 3 2 θ+8 θ+3 3 =0, θ ∈[-3 π, 2 π] is: JEE Main 2026 The number of solutions of the equation 2 θ (θ)/(2)+ (5 θ)/(2)=2 ^3 (5 θ)/(2) in… JEE Main 2025 If for θ ∈[-(π)/(3), 0] , the points (x, y)=(3 (θ+(π)/(3)), 2 (θ+(π)/(6))) lie on x y+α… JEE Main 2025 If 10 ^4 θ+15 ^4 θ=6 , then the value of 27 cosec^6 θ+8 ^6 θ 16 ^8 θ is: JEE Main 2025 The number of solutions of equation (4- 3 ) x -2 3 ^2 x=- 4 1+ 3 , x ∈[-2 π, (5 π)/(2)] is JEE Main 2025 The number of solutions of the equation 2 x+3 x=π, x ∈[-2 π, 2 π]-\ ± (π)/(2), ± (3… JEE Main 2025 If θ ∈[-(7 π)/(6), (4 π)/(3)] , then the number of solutions of 3 cosec^2 θ-2( 3 -1)… JEE Main 2025 If θ ∈[-2 π, 2 π] , then the number of solutions of 2 2 ^2 θ+(2- 6 ) θ- 3 =0 , is equal… JEE Main 2025 The sum of all values of θ ∈[0,2 π] satisfying 2 ^2 θ= 2 θ and 2 ^2 θ=3 θ is JEE Main 2025 Let f: ℝ arrow ℝ be a function defined by f(x)=\ array cc x^2 ((π)/(x^2)), & if x ≠ 0 0… JEE Advanced 2024 Full Trigonometric Equations list Trigonometric Equations JEE Maths PYQs JEE Mathematics PYQs All NTA Abhyas JEE Main PYQs