Question
Prove the following trigonometric identities.
$\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}=\cot\text{A}\tan\text{B}$
$\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}=\cot\text{A}\tan\text{B}$
✓
Answer
We have to prove $\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}=\cot\text{A}\tan\text{B}$
Now, $\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}=\frac{\cot\text{A}+\frac{1}{\cot\text{B}}}{\cot\text{B}+\frac{1}{\cot\text{A}}}$
$\text{L.H.S}=\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}$
$=\frac{\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{B}}}{\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{B}}}$
$=\frac{\cot\text{A}}{\cot\text{B}}$
$=\cot\text{A}\frac{1}{\cot\text{B}}$
$=\cot\text{A}\tan\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Now, $\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}=\frac{\cot\text{A}+\frac{1}{\cot\text{B}}}{\cot\text{B}+\frac{1}{\cot\text{A}}}$
$\text{L.H.S}=\frac{\cot\text{A}+\tan\text{B}}{\cot\text{B}+\tan\text{A}}$
$=\frac{\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{B}}}{\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{B}}}$
$=\frac{\cot\text{A}}{\cot\text{B}}$
$=\cot\text{A}\frac{1}{\cot\text{B}}$
$=\cot\text{A}\tan\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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