MCQ
$\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$ is equal to :
- A$0$
- B$1$
- ✓$\sin\theta+\cos\theta$
- D$\sin\theta-\cos\theta$
✓
Answer
Correct option: C.
$\sin\theta+\cos\theta$
$\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}=\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}$
$=\frac{\sin\theta\times\sin\theta}{\sin\theta-\cos\theta}+\frac{\cos\theta\times\cos\theta}{\cos\theta-\sin\theta}$
$=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}$
$=\sin\theta+\cos\theta$
$=\frac{\sin\theta\times\sin\theta}{\sin\theta-\cos\theta}+\frac{\cos\theta\times\cos\theta}{\cos\theta-\sin\theta}$
$=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}$
$=\sin\theta+\cos\theta$
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|
Class
|
13.8-14
|
14-14.2
|
14.2-14.4
|
14.4-14.6
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14.6-14.8
|
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