Question

prove that cos⁡ 2B - cos ⁡2A / sin⁡ 2A + sin ⁡2B = tan (A-B)

Answer 1

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Answer 2

Answer and explanation:

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

Proof :

Taking LHS,

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

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

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

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

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

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

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

$=\tan(A-B)$

=RHS