Prove that
Question:

Prove that

$\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$

 

Solution:

$=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$

$=\frac{2 \sin \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}+2 \sin \frac{9 x+3 x}{2} \cos \frac{9 x-3 x}{2}}{2 \cos \frac{7 x^{2}+5 x}{2} \cos \frac{7 x^{2}-5 x}{2}+2 \cos \frac{9 x^{2}+3 x}{2} \cos \frac{9 x^{2}-3 x}{2}}$

$=\frac{2 \sin 6 x \cos x+2 \sin 6 x \cos 3 x}{2 \cos 6 x \cos x+2 \cos 6 x \cos 3 x}$

$=\frac{2 \sin 6 x(\cos x+\cos 3 x)}{2 \cos 6 x(\cos x+\cos 3 x)}$

$=\frac{\sin 6 x}{\cos 6 x}$

$=\tan 6 x$

Using the formula,

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

$\cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

 

 

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