SATHEE: Quantitative Aptitude Ques 1726

Quantitative Aptitude Ques 1726

Question: If $ (sinx+\sin y)=a $ and $ (\cos x+\cos y)=b, $ what is the value of $ \sin x\sin y+\cos xcosy? $

Options:

A) $ a+b-ab $

B) $ a+b+ab $

C) $ a^{2}+b^{2}-2 $

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

E) None of these

Show Answer

Answer:

Correct Answer: D

Solution:

$ (\sin x+\sin y)=a $ and $ (\cos x+\cos y)=b $ On squaring both the equations, we get $ {{(\sin x+\sin y)}^{2}}=a^{2} $ $ {{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y=a^{2} $ (i) and $ {{(\cos x+\cos y)}^{2}}=b^{2} $ $ {{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y=b^{2} $ … (ii) On adding Eqs. (i) and (ii), we get $ ({{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y) $ $ +({{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y)=a^{2}+b^{2} $

$ \Rightarrow $ $ {{\sin }^{2}}x+{{\cos }^{2}}x+{{\sin }^{2}}y+{{\cos }^{2}}y $ $ +2(\sin x\sin y+\cos x\cos y)=a^{2}+b^{2} $

$ \Rightarrow $ $ 1+1+2(\sin x\sin y+\cos x\cos y)=a^{2}+b^{2} $

$ \therefore $ $ \sin x\sin y+\cos x\cos y=\frac{a^{2}+b^{2}-2}{2} $

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