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Manipal Entrance Test 2014 Math Paper

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Question : 11 of 80

Marks: +1, -0

If

\tan \frac{\alpha}{2},

\tan \frac{\beta}{2}

are the rootsof

8 x^{2} - 26 x + 15 = 0,

\cos (\alpha + \beta)

is equal to

$\frac{627}{725}$
$-\frac{627}{725}$
-1
None of these

Solution:

The given equation is

8x^{2} -2 6 x + 15 = 0.

∴ The sum of roots,

\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} = \frac{26}{8} = \frac{13}{4}

and product of roots,

\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} = \frac{15}{8}

∴

\tan \left( \frac{\alpha + \beta}{2} \right)

= \frac{ \tan \frac{\alpha}{2} + \tan \frac{\beta}{2} }{ 1 - \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} }

= \frac{ \frac{13}{4} }{ 1 - \frac{15}{8} } = -\frac{26}{7}

Now,

\cos (\alpha + \beta) = \frac{ 1 - \tan^{2} \left( \frac{\alpha + \beta}{2} \right) }{ 1 + \tan^{2} \left( \frac{\alpha + \beta}{2} \right) }

\left[ \because \cos 2\theta = \frac{ 1 - \tan^{2} \theta }{ 1 + \tan^{2} \theta } \right]

= \frac{ 1 - \left( -\frac{26}{7} \right)^{2} }{ 1 + \left( -\frac{26}{7} \right)^{2} } = \frac{ 49 - 676 }{ 49 + 676 }

= -\frac{627}{725}

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