A) \[\sin \frac{A-B}{2}\]
B) \[\tan \frac{A-B}{2}\]
C) \[\cot \frac{A-B}{2}\]
D) None of these
Correct Answer: B
Solution :
We have \[\sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}\] \[\Rightarrow \frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}\] \[=\tan \frac{A-B}{2}\cot \frac{A+B}{2}\] \[\Rightarrow \frac{n-1}{n+1}\tan \left( \frac{A+B}{2} \right)=\tan \frac{A-B}{2}\] .
You need to login to perform this action.
You will be redirected in
3 sec
