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sin θ1−cot θ+cos θ1−tan θ is equal to

Question:

$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ is equal to

(a) 0

(b) 1

(c) $\sin \theta+\cos \theta$

(d) $\sin \theta-\cos \theta$

Solution:

The given expression is $\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$.

Simplifying the given expression, we have

$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$

$=\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}$

$=\frac{\sin \theta}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\cos \theta}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$

$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}+\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}+\frac{\cos ^{2} \theta}{-1(\sin \theta-\cos \theta)}$

$=\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}-\frac{\cos ^{2} \theta}{\sin \theta-\cos \theta}$

$=\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin \theta-\cos \theta}$

$=\frac{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}{(\sin \theta-\cos \theta)}$

$=\sin \theta+\cos \theta$

Therefore, the correct option is (c).

 

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