If sin A = n sin B, then frac{n - 1}{n + 1} t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$Applying the componendo and dividendo rule,we get $\frac{n - 1}{n + 1} =

Vedclass · Accepted Answer

$	an \frac{A - B}{2}$

Vedclass · Answer

(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$Applying the componendo and dividendo rule,we get $\frac{n - 1}{n + 1} = \frac{\sin A - \sin B}{\sin A + \sin B}.$Using the sum-to-product formulas $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$ and $\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2},$$\frac{n - 1}{n + 1} = \frac{2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \cot \frac{A + B}{2} 	an \frac{A - B}{2}.$Multiplying both sides by $	an \frac{A + B}{2},$ we get $\frac{n - 1}{n + 1} 	an \frac{A + B}{2} = 	an \frac{A - B}{2} \cot \frac{A + B}{2} 	an \frac{A + B}{2} = 	an \frac{A - B}{2}.$

If $\sin A = n \sin B,$ then $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = $

Similar Questions

$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} = $

The exact value of $\cos \frac{2\pi}{28} \csc \frac{3\pi}{28} + \cos \frac{6\pi}{28} \csc \frac{9\pi}{28} + \cos \frac{18\pi}{28} \csc \frac{27\pi}{28}$ is equal to

$\frac{\cos A}{1 - \sin A} = $

If $\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$,where $P$ and $Q$ both are acute angles,then the value of $P - Q$ is....$^o$

The value of $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module