Trigonometric Identities Question 170
Question: If $ \tan \theta =\frac{x,\sin ,\varphi }{1-x,\cos ,\varphi } $ and $ \tan ,\varphi =\frac{y\sin ,\theta }{1-y,\cos ,\theta } $ , then $ \frac{x}{y}= $
[MP PET 1991]
Options:
A) $ \frac{\sin \varphi }{\sin \theta } $
B) $ \frac{\sin \theta }{\sin \varphi } $
C) $ \frac{\sin \varphi }{1-\cos \theta } $
D) $ \frac{\sin \theta }{1-\cos \varphi } $
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Answer:
Correct Answer: B
Solution:
We have $ \tan \theta =\frac{x,\sin ,\varphi }{1-x,\cos \varphi } $
$ \Rightarrow \frac{1}{x}\tan \theta -\tan \theta \cos \varphi =\sin ,\varphi $
$ \Rightarrow \frac{1}{x}=\frac{\sin ,\varphi +\cos \varphi ,\tan ,\theta }{\tan ,\theta } $ and $ \tan ,\varphi =\frac{y,\sin ,\theta }{1-y,\cos ,\theta } $
$ \Rightarrow \tan ,\varphi ,=\frac{\sin ,\theta }{\frac{1}{y}-\cos ,\theta } $
$ \Rightarrow \frac{1}{y}\tan ,\varphi -\tan ,\varphi \cos \theta =\sin \theta $
$ \Rightarrow \frac{1}{y}\tan ,\varphi ,=\sin ,\theta +\tan ,\varphi ,\cos \theta $
$ \therefore \frac{1}{y}=\frac{\sin ,\theta +\tan ,\varphi ,\cos \theta }{\tan ,\varphi } $ Now $ \frac{x}{y}=[ \frac{\tan ,\theta }{\sin ,\varphi +\cos ,\varphi ,\tan ,\theta } ]\times [ \frac{\sin ,\theta +\tan ,\varphi ,\cos ,\theta }{\tan ,\varphi } ] $ $ =\frac{\tan ,\theta }{\tan ,\varphi },[ \frac{\sin ,\theta +\cos ,\theta \frac{\sin \varphi }{\cos \varphi }}{\sin \varphi +\cos \varphi \frac{\sin \theta }{\cos \theta }} ]=\frac{\tan \theta \cos \theta }{\tan ,\varphi ,\cos \varphi }=\frac{\sin \theta }{\sin \varphi } $ Aliter: $ x,\sin ,\varphi =\tan ,\theta -x,\cos ,\varphi ,\tan ,\theta $
$ \Rightarrow ,x=\frac{\tan ,\theta }{\sin ,\varphi +\cos ,\varphi ,\tan ,\theta } $ $ =\frac{\sin ,\theta }{\cos ,\theta \sin ,\varphi +\cos ,\varphi ,\sin ,\theta }=\frac{\sin ,\theta }{\sin ,(\theta +\varphi )} $ Similarly, $ y=\frac{\sin ,\varphi }{\sin ,(\theta +\varphi )} $ ;
$ \therefore \frac{x}{y}=\frac{\sin \theta }{\sin \varphi }. $
BETA
