SATHEE: Quantitative Aptitude Ques 2391

Quantitative Aptitude Ques 2391

Question: If $ \frac{\cos ,x}{\cos y}=n $ and $ \frac{\sin ,x}{\sin y}=m, $ then what is the value of $ (m^{2}-n^{2}){{\sin }^{2}}y $

Options:

A) $ 1-n^{2} $

B) $ 1+n^{2} $

C) $ m^{2} $

D) $ n^{2} $

Show Answer

Answer:

Correct Answer: A

Solution:

$ (m^{2}-n^{2}){{\sin }^{2}}y=( \frac{{{\sin }^{2}}x}{{{\sin }^{2}}y}-\frac{{{\cos }^{2}}x}{{{\cos }^{2}}y} ){{\sin }^{2}}y $ $ =\frac{{{\sin }^{2}}x{{\cos }^{2}}y-{{\cos }^{2}}x{{\sin }^{2}}y}{{{\sin }^{2}}y{{\cos }^{2}}y}\times {{\sin }^{2}}y $ $ =,\frac{{{\sin }^{2}}x,(1-{{\sin }^{2}})-(1-{{\sin }^{2}}x)\times {{\sin }^{2}}y}{{{\cos }^{2}}y} $ $ =,\frac{{{\sin }^{2}}x-{{\sin }^{2}}x{{\sin }^{2}}y-{{\sin }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y}{{{\cos }^{2}}y} $ $ =\frac{1-{{\cos }^{2}}x-1+{{\cos }^{2}}y}{{{\cos }^{2}}y} $ $ =\frac{{{\cos }^{2}}y-{{\cos }^{2}}x}{{{\cos }^{2}}y}=1-\frac{{{\cos }^{2}}x}{{{\cos }^{2}}y}=1-n^{2} $

Quantitative Aptitude Ques 2390

Quantitative Aptitude Ques 2392

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