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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Let $A$ and $B$ denote the statements $A=\cos \alpha+\cos \beta+\cos\gamma=0,B=\sin\alpha+\sin\beta+\sin\gamma=0$.If $\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\large\frac{3}{2}$ then

$\begin{array}{1 1}(a)\;A\;is \;false\;and \;B\;is \;true\\(b)\;Both\; A \;and \;B\;are\;true\\(c)\;Both\;A\;and \;B\;are\;false\\(d)\;A\;is\; true\;and\;B\;is \;false\end{array}$

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1 Answer

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$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\large\frac{3}{2}$
$\Rightarrow 2[\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)]+3=0$
$\Rightarrow 2[\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)]+\sin^2\alpha+\cos^2\alpha+\sin^2\beta+\cos^2\beta+\sin^2\gamma+\cos^2\alpha=0$
$\Rightarrow [\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta+2\sin\beta\sin\gamma+2\sin\gamma\sin\alpha]+[\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta+2\cos\beta\cos\gamma+2\cos\gamma\cos\alpha]=0$
$\Rightarrow [\sin\alpha+\sin\beta+\sin\gamma]^2+[\cos\alpha+\cos\beta+\cos\gamma]^2=0$
$\sin\alpha+\sin\beta+\sin\gamma=0$
$\cos\alpha+\cos\beta+\cos\gamma=0$
$\therefore $ A and B both are true.
Hence (b) is the correct answer.
answered Oct 17, 2013 by sreemathi.v
 

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