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What is $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$ equal to?
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2836 Upvotes
Verified Answer
The correct answer is:
$\sin \theta+\cos \theta$
$$
\text { } \begin{aligned}
& \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \\
& =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin ^2 \theta}{\sin \theta-\cos \theta} \\
& =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta-\sin \theta} \\
& =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta-\sin \theta} \\
& =\frac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}{(\cos \theta-\sin \theta)} \\
& {\left[\text { Using identity } a^2-b^2=(a+b)(a-b)\right]} \\
& =\sin \theta+\cos \theta
\end{aligned}
$$
\text { } \begin{aligned}
& \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)} \\
& {\left[\text { Using identity } a^2-b^2=(a+b)(a-b)\right]} \\
& =\sin \theta+\cos \theta
\end{aligned}
$$
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