Question

If $\sin A=\frac{3}{5},0<A<\frac{\pi }{2}$ and $\cos B=\frac{-12}{13},\pi <B<\frac{3\pi }{2}$, then value of $\sin (A-B)$ is

• A. $-\frac{13}{82}$
• B. $-\frac{15}{65}$
• C. $-\frac{13}{75}$
• D. $-\frac{16}{65}$

Answer 1

Answer: (D)

We have $:\sin A=\frac{3}{5}$, where $0<A<\frac{\pi }{2}$
$\therefore \cos A=\pm \sqrt{1-{\sin }^{2}A}$
$\Rightarrow \cos A=+\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$[\because \cos$ is positive in first quadrant $]$
It is given that $:\cos B=\frac{-12}{13}$ and $\pi <B<\frac{3\pi }{2}$
$\therefore \sin B=\pm \sqrt{1-{\cos }^{2}B}\Rightarrow \sin B=-\sqrt{1-{\cos }^{2}B}$
$[\because$ Sine is negative in the third quadrant]
$\Rightarrow \sin B=-\sqrt{1-{\left(\frac{-12}{13}\right)}^2}=-\frac{5}{13}$
Now, $\sin (A-B)=\sin A\cos B-\cos A\sin B$
$=\frac{3}{5}\times \frac{-12}{13}-\frac{4}{5}\times \frac{-5}{13}=-\frac{16}{65}$