Question

If $0<\text{A}<\text{B}<\pi ,sinA-sinB=\frac{1}{\sqrt{2}}$ and $cosA-cosB=\sqrt{\frac{3}{2}}$ , 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)

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