Question

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

• A. $\frac{2 \pi }{3}$
• B. $\frac{5 \pi }{6}$
• C. $\pi $
• D. $\frac{4 \pi }{3}$

Answer 1

Answer: (D)

$\left(sin A - sin ⁡ B\right)^{2}+\left(cos ⁡ A - cos ⁡ B\right)^{2}=\frac{1}{2}+\frac{3}{2}=2$
$\Rightarrow 2-2\left(sin A sin ⁡ B + cos ⁡ A cos ⁡ B\right)=2$
$\Rightarrow cos\left(B - A\right)=0\Rightarrow B-A=\frac{\pi }{2}\Rightarrow B=A+\frac{\pi }{2}$
$sin A-sin ⁡ B=\frac{1}{\sqrt{2}}\Rightarrow sin ⁡ A-cos ⁡ A=\frac{1}{\sqrt{2}}$
$\Rightarrow sin A.\frac{1}{\sqrt{2}}-cos ⁡ A.\frac{1}{\sqrt{2}}=\frac{1}{2}$
$\Rightarrow sin\left(A - \frac{\pi }{4}\right)=sin\frac{\pi }{6}$
$\Rightarrow A=\frac{\pi }{6}+\frac{\pi }{4}=\frac{5 \pi }{12}$
$A+B=A+A+\frac{\pi }{2}=\frac{5 \pi }{6}+\frac{\pi }{2}=\frac{4 \pi }{3}$