Prove that: (A+B) (A-B) = ^2A- ^2B 1- ^2A ^2B - Vidyadip

Question

Prove that: $\frac{\tan\text{(A}+\text{B)}}{\cot\text{(A}-\text{B)}}=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-\tan^2\text{A}\tan^2\text{B}}$

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Answer

$\text{L.H.S}=\frac{\tan\text{(A}+\text{B)}}{\cot\text{(A}-\text{B)}}$ $=\frac{\frac{\tan\text{(A}+\text{B)}}{1}}{\tan\text{(A}-\text{B)}}$ $=\tan\text{(A}+\text{B)}\tan\text{(A}-\text{B)}$ $=\Big[\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]\Big[\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}\Big]$ $=\frac{(\tan\text{A}+\tan\text{B)}(\tan\text{A}-\tan\text{B)}}{(1-\tan\text{A}\tan\text{B)}(1+\tan\text{A}\tan\text{B)}}$ $=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-(\tan\text{A}\tan\text{B)}^2}$ $=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-\tan^2\text{A}\tan^2\text{B}}$ $\big[\because\text{(a}-\text{b)}\text{(a}+\text{b)}=\text{a}^2-\text{b}^2\big]$ $=\text{R.H.S}$ $\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.

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