Question

निम्नलिखित को सिद्ध कीजिए: $\dfrac {\cos 9x - \cos 5x}{ \sin 17x - \sin 3x} = - \dfrac{\sin 2x}{ \cos 10x}$

Answer 1

Answer:

Step-by-step explanation:

cos A-cos B=-2 sin ((A+B)/2) sin ((A-B)/2)

cos9x-cos 5x=-2 sin((9x+5x)/2) sin((9x-5x)/2)=-2 sin 7x sin2x

sin A -sin B=2 cos ((A+B)/2) sin ((A-B)/2)

sin 17x-sin 3x=2 cos ((17x+3x)/2) sin ((17x-3x)/2)=2 cos 10x sin 7x

Now

cos9x-cos 5x/sin 17x-sin 3x = -2 sin 7x sin2x/2 cos 10x sin 7x

cancel sin7x and 2

Now it becomes

cos9x-cos 5x/sin 17x-sin 3x= - sin2x/cos10x

Answer 2

निम्नलिखित को सिद्ध कीजिए: $\dfrac {\cos 9x - \cos 5x}{ \sin 17x - \sin 3x} = - \dfrac{\sin 2x}{ \cos 10x}$

जैसा कि हम जानते हैं;

$sin \: A -sin \: B = 2cos( \frac{A+B}{2} ) sin( \frac{A-B}{2}) \\ \\ cos \: A- cos \: B = 2sin( \frac{A+B}{2} ) sin( \frac{B-A}{2})$

$\dfrac {\cos 9x - \cos 5x}{ \sin 17x - \sin 3x} = - \dfrac{\sin 2x}{ \cos 10x}$

$\dfrac { 2sin( \frac{9x + 5x}{2} ) sin( \frac{5x - 9x}{2}) }{ 2cos( \frac{17x + 3x}{2} ) sin( \frac{17x - 3x}{2})} \\ \\ = > \frac{2sin \: 7x \: sin( - 2x)}{2cos \: 10x \: sin(7x)} \\ \\ = > \frac{sin( - 2x)}{cos \: 10x} \\ \\ = > - \frac{sin(2x)}{cos \: 10x} \\ \\ = > RHS$

क्योंकि $sin(-x)=-sin x$