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What is $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$ equal to?
Options:
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Verified Answer
The correct answer is:
$\sin \theta+\cos \theta$
$\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$
$=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}+\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta-\sin \theta}=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}$
$=\cos \theta+\sin \theta$
$\therefore \quad$ Option (b) is correct.
$=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}+\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta-\sin \theta}=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}$
$=\cos \theta+\sin \theta$
$\therefore \quad$ Option (b) is correct.
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