If $\tan (A+B)=n \tan (A-B)$ show that : $ (n+1) \sin 2 B=(n-1) \sin 2 A $
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Answer
Given, $\frac{\tan ( A + B )}{\tan ( A - B )}=\frac{n}{1}$ From Compondendo-Dividendo Rule $\begin{array}{l}\Rightarrow \quad \frac{\tan (A+B)+\tan (A-B)}{\tan (A+B)-\tan (A-B)}=\frac{n+1}{n-1} \\\Rightarrow \quad \frac{\frac{\sin (A+B)}{\cos (A+B)}+\frac{\sin (A-B)}{\cos (A-B)}}{\frac{\sin (A+B)}{\cos (A+B)}-\frac{\sin (A-B)}{\cos (A-B)}}=\frac{n+1}{n-1}\end{array}$ $\Rightarrow \frac{\frac{\sin ( A + B ) \cos ( A - B )+\sin ( A - B ) \cos ( A + B )}{\cos ( A + B ) \cos ( A - B )}}{\frac{\sin ( A + B ) \cos ( A - B )-\sin ( A - B ) \cos ( A + B )}{\cos ( A + B ) \cos ( A - B )}}=\frac{n+1}{n-1}$ $\Rightarrow \quad \frac{\sin [( A + B )+( A - B )]}{\sin [( A + B )-( A - B )]}=\frac{n+1}{n-1}$ $\Rightarrow \quad \frac{\sin 2 A}{\sin 2 B}=\frac{n+1}{n-1}$ $\Rightarrow(n+1) \sin 2 A=(n-1) \sin 2 A$ Hence proved.
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