If frac{sin(x + y)}{sin(x - y)} = frac{a + b} | vedclass

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

Question

(B) Given: $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$Applying the Componendo and Dividendo rule,which states that if $\frac{p}{q} = \frac

Vedclass · Accepted Answer

$\frac{a}{b}$

Vedclass · Answer

(B) Given: $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$Applying the Componendo and Dividendo rule,which states that if $\frac{p}{q} = \frac{r}{s},$ then $\frac{p + q}{p - q} = \frac{r + s}{r - s}$:$\frac{\sin(x + y) + \sin(x - y)}{\sin(x + y) - \sin(x - y)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)}$Using the trigonometric identities $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$ and $\sin(A + B) - \sin(A - B) = 2\cos A \sin B$:$\frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}$$\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$$	an x \cdot \frac{1}{	an y} = \frac{a}{b}$$\frac{	an x}{	an y} = \frac{a}{b}$

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

Similar Questions

Which of the following relations is correct for $0^{\circ} < \theta < 90^{\circ}$?

The value of $\cos (270^\circ + \theta )\,\cos (90^\circ - \theta ) - \sin (270^\circ - \theta )\,\cos \theta $ is

Let $f(x) = \sin^2 x + \cos^4 x + 2$ and $g(x) = \cos(\cos x) + \cos(\sin x)$. If the periods of $f(x)$ and $g(x)$ are $T_1$ and $T_2$ respectively,then:

The value of $\csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18}$ is a

If $f(x) = \cos^2 x + \sec^2 x$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module