A) \[x=y\]
B) \[x+y=2\]
C) \[x=2y\]
D) \[y=2x\]
Correct Answer: C
Solution :
[c]| We have, \[x=\tan 27\theta -\tan \theta \] |
| \[y=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }\] |
| \[\because \]\[\tan A-\tan B=\frac{\sin (A-B)}{\cos A\cos B}\] |
| \[\therefore \]\[\tan 3\theta -\tan \theta =\frac{\sin 2\theta }{\cos 3\theta \cos \theta }\]\[=\frac{2\sin \theta \cos \theta }{\cos 3\theta \cos \theta }=\frac{2\sin \theta }{\cos 3\theta }\] |
| \[\therefore \]\[\frac{\sin \theta }{\cos 3\theta }=\frac{1}{2}(\tan 3\theta -\tan \theta )\] |
| \[\therefore \]\[y=\frac{1}{2}\] \[[\tan 3\theta -\tan \theta +\tan 9\theta -\tan 3\theta +\tan 27\theta -tan9\theta ]\] |
| \[\Rightarrow \]\[y=\frac{1}{2}(\tan 27-\tan \theta )\] |
| \[\Rightarrow \]\[2y=x\]\[\Rightarrow \]\[x=2y\] |
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