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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| $(i)$ | $((\text{A}'\cup\text{B}')-\text{A})'$ | $(a)$ | $\text{A} - \text{B}$ |
| $(ii)$ | $[\text{B}'\cup(\text{B}'-\text{A})]'$ | $(b)$ | $\text{A}$ |
| $(iii)$ | $(\text{A} - \text{B}) - (\text{B} - \text{C})$ | $(c)$ | $\text{B}$ |
| $(iv)$ | $(\text{A}-\text{B})\cap(\text{C}-\text{B})$ | $(d)$ | $(\text{A}\times\text{B})\cap(\text{A}\times\text{C})$ |
| $(v)$ | $\text{A}\times(\text{B}\cap\text{C})$ | $(e)$ | $(\text{A}\times\text{B})\cup(\text{A}\times\text{C})$ |
| $(vi)$ | $\text{A}\times(\text{B}\cup\text{C})$ | $(f)$ | $(\text{A}\cap\text{C})-\text{B}$ |