Question

If y sin φ = x sin ( 2 θ + φ ) .

Then show that :

( x + y ) cot ( θ + φ ) = ( y . x ) cot θ

Answer 1

Correct Questions:

If y sin φ = x sin ( 2 θ + φ ) .

Then show that :

( x + y ) cot ( θ + φ ) = ( y - x ) cot φ

Proof:

For convenience, let's denote

φ as ß and θ as α

Given,

$y \sin( \beta ) = x \sin(2 \alpha + \beta )$

$= > \frac{x}{y} = \frac{ \sin( \ \beta ) }{ \sin(2 \alpha + \beta ) }$

Applying Componendo and dividendo,

we get,

$= > \frac{x + y}{x - y} = \frac{ \sin( \beta ) + \sin(2 \alpha + \beta ) }{ \sin( \beta ) - \sin(2 \alpha + \beta ) }$

$= \frac{2 \sin( \frac{2 \alpha + 2 \beta }{2} ) \cos (\frac{ \beta - 2 \alpha - \beta }{2} ) }{2 \cos (\frac{2 \alpha + 2 \beta }{2} ) \sin( \frac{ \beta - 2 \alpha - \beta }{2} ) } \\ \\ = - \frac{ \cot( \beta ) }{ \cot( \alpha + \beta ) }$

Therefore,

$= > \frac{x + y}{x - y} = - \frac{ \cot( \beta ) }{ \cot( \alpha + \beta ) } \\ \\ = > {(x + y)}{ \cot( \alpha + \beta ) } = (y - x) \cot( \beta )$

=> ( x + y ) cot ( θ + φ ) = ( y - x ) cot φ

Hence, proved

$\bold\red{Concepts\:Used:}$

1. By componendo and Dividendo,

if, $\frac{a}{b} = \frac{x}{y}$

then, $\frac{a + b}{a - b} = \frac{x + y}{x - y}$

2. $\sin( \alpha ) + \sin( \beta ) = 2 \sin( \frac{ \alpha + \beta }{2} ) \cos( \frac{ \alpha - \beta }{2} )$

3. $\sin( \alpha ) - \sin( \beta ) = 2 \cos( \frac{ \alpha + \beta }{2} ) \sin( \frac{ \alpha - \beta }{2} )$

Answer 2

Step- by - step explanation:-

According to the question→

$y \: sin \phi \: = x \: sin \: (2 \theta + \phi) \\ \\ \frac{y}{x} = \frac{sin \: (2 \theta + \phi)}{sin \phi}$

using componendo and dividendo rule

Is →

$\implies \: \frac{x}{y} = \frac{a}{b} \\ \implies \: \frac{x + y}{x - y} = \frac{a + b}{a - b}$

$\frac{y + x}{y - x} = \frac{sin(2 \theta + \phi) + sin \phi}{sin(2 \theta + \phi) - sin \phi}$

Using formula→

$\star \: sin \: c \: + sin \: d = 2sin \frac{c + d}{2} cos \: \frac{c - d}{2} \\ \\ \star \: sin \: c \: - sin \: d = 2cos \: \frac{c + d}{2} sin \frac{c - d}{2} \\ \\ \therefore \\ \\ \frac{y + x}{y - x} = \frac{2sin( \theta + \phi).cos \theta}{2cos( \theta + \phi).sin \phi} \\ \\ \frac{y + x}{y - x} = \frac{tan ( \theta + \phi)}{tan \phi} \\ \\ (x + y).tan \phi = (y - x)tan( \theta + \phi) \\ \\ \boxed{(x + y)cot ( \theta + \phi) = (y - x)cot \phi}$

Hence proved.