SATHEE: Quantitative Aptitude Ques 565

Quantitative Aptitude Ques 565

Question: If $ (a^{2}-b^{2})\sin \theta +2ab\cos \theta =a^{2}+b^{2}, $ then the value of tan $ \theta $ is

Options:

A) $ \frac{1}{2ab},(a^{2}+b^{2}) $

B) $ \frac{1}{2}(a^{2}-b^{2}) $

C) $ \frac{1}{2ab},(a^{2}-b^{2}) $

D) $ \frac{1}{2}(a^{2}+b^{2}) $

Show Answer

Answer:

Correct Answer: C

Solution:

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

$ \therefore $ $ \sin \alpha sin\theta +\cos \alpha \cos \theta =1 $

$ \Rightarrow $ $ \cos ,(\alpha -\theta )=1=cos,(0{}^\circ ) $

$ \Rightarrow $ $ \alpha -\theta =0{}^\circ $
$ \Rightarrow $ $ \alpha =\theta $

$ \Rightarrow $ $ \tan \alpha =\tan \theta =\frac{a^{2}-b^{2}}{2ab} $

Quantitative Aptitude Ques 564

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