Question
Prove the following trigonometric 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
$\text{L.H.S}=\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{B}}{\sin\text{B}}}$
$=\frac{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{(\cos\text{A}\cos\text{B})}}{\frac{\cos\text{A}\sin\text{B}+\sin\text{A}\cos\text{B}}{\sin\text{A}\sin\text{B}}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}\times\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}}$
$=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\tan\text{A}\cdot\tan\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{B}}{\sin\text{B}}}$
$=\frac{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{(\cos\text{A}\cos\text{B})}}{\frac{\cos\text{A}\sin\text{B}+\sin\text{A}\cos\text{B}}{\sin\text{A}\sin\text{B}}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}\times\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}}$
$=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\tan\text{A}\cdot\tan\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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