Question
Prove the following identities:
$\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}=\tan\text{A}\tan\text{B}$
$\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}=\tan\text{A}\tan\text{B}$
✓
Answer
$\text{LHS}=\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{B}}{\sin\text{ B}}}$
$=\frac{\frac{\sin\text{B}\cos\text{B}+\sin\text{B}\cos\text{A}}{\cos\text{A}\cos\text{B}}}{\frac{\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}}$
$=\frac{(\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A})\times\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}\times\big(\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}\big)}$
$=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\tan\text{A}\tan\text{B}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{B}}{\sin\text{ B}}}$
$=\frac{\frac{\sin\text{B}\cos\text{B}+\sin\text{B}\cos\text{A}}{\cos\text{A}\cos\text{B}}}{\frac{\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}}$
$=\frac{(\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A})\times\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}\times\big(\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}\big)}$
$=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\tan\text{A}\tan\text{B}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
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