Question
Prove the following trigonometric identities.
$\frac{\cos\text{A}}{1-\tan\text{A}}+\frac{\sin\text{A}}{1-\cot\text{A}}=\sin\text{A}+\cos\text{A}$

Answer

$\text{L.H.S}=\frac{\cos\text{A}}{1-\tan\text{A}}+\frac{\sin\text{A}}{\Big(1-\frac{1}{\tan\text{A}}\Big)}$
$=\frac{\cos\text{A}}{1-\tan\text{A}}-\frac{\sin\text{A}.\tan\text{A}}{1-\tan\text{A}}$
$\Rightarrow\ \frac{\cos\text{A}-\sin\text{A}\tan\text{A}}{(1-\tan\text{A})}$
$\Rightarrow\ \frac{\cos\text{A}-\sin\text{A}\times\frac{\sin\text{A}}{\cos\text{A}}}{1-\frac{\sin\text{A}}{\cos\text{A}}}$
$\Rightarrow\ \frac{\cos^2\text{A}-\sin^2\text{A}\cos\text{A}}{(\cos\text{A}-\sin\text{A})\cos\text{A}}$
$=\frac{(\cos\text{A}-\sin\text{A})(\cos\text{A}+\sin\text{A})}{\cos\text{A}-\sin\text{A}}$
$\Rightarrow\ \cos\text{A}+\sin\text{A}=\text{R.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$

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