If 0<A<B<π, sin⁡A-sin⁡B=1√2 and cos⁡A-cos⁡B=√32, 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 ⇒2-2sin⁡Asin⁡B+cos⁡Acos⁡B=2 ⇒cosB-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 ⇒sinA-(π)/4=sin(π)/6 ⇒A=(π)/6+(π)/4=(5π)/12 A+B=A+A+(π)/2=(5π)/6+(π)/2=(4π)/3

Q37 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

If $0 < A < B < π, \sin A - \sin B = \frac{1}{\sqrt{2}}$ and $\cos A - \cos B = \sqrt{\frac{3}{2}}$ , then the value of $A + B$ is equal to

A
$\frac{2 π}{3}$
B
$\frac{5 π}{6}$
C
$π$
Correct answer
D
$\frac{4 π}{3}$

Solution

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

\Rightarrow 2 - 2 (\sin A \sin B + \cos A \cos B) = 2

\Rightarrow c o s (B - A) = 0 \Rightarrow B - A = \frac{π}{2} \Rightarrow B = A + \frac{π}{2}

\sin A - \sin B = \frac{1}{\sqrt{2}} \Rightarrow \sin A - \cos A = \frac{1}{\sqrt{2}}

\Rightarrow \sin A . \frac{1}{\sqrt{2}} - \cos A . \frac{1}{\sqrt{2}} = \frac{1}{2}

\Rightarrow s i n (A - \frac{π}{4}) = s i n \frac{π}{6}

\Rightarrow A = \frac{π}{6} + \frac{π}{4} = \frac{5 π}{12}

A + B = A + A + \frac{π}{2} = \frac{5 π}{6} + \frac{π}{2} = \frac{4 π}{3}

Hence, the correct answer is option: 'D'.

If 0<A<B<π, sin⁡A-sin⁡B=12 and cos⁡A-cos⁡B=32, then the value of A+B is equal to

Solution

More Trigonometric Equations Questions

If $0 < A < B < π, \sin A - \sin B = \frac{1}{\sqrt{2}}$ and $\cos A - \cos B = \sqrt{\frac{3}{2}}$ , then the value of $A + B$ is equal to