Question

If \[\tan \left( \dfrac{\alpha }{2} \right)\] and \[\tan \left( \dfrac{\beta }{2} \right)\] are the roots of \[8{{x}^{2}}-26x+15=0\], then \[\cos \left( \alpha +\beta \right)\] is equal to
a) \[\dfrac{627}{726}\]
b) \[-\left( \dfrac{627}{725} \right)\]
c) \[-1\]
d) None of these.

Answer 1

Hint: We know that if \[\alpha \] and \[\beta \] are the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] then \[\alpha +\beta =-\dfrac{b}{a}\] and \[\alpha \times \beta =\dfrac{c}{a}\]. Using this concept and some trigonometric formula we can solve the given question.

Formula Used:
i) \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
ii) \[\cos 2A=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]

Complete step by step solution:
Given \[\tan \left( \dfrac{\alpha }{2} \right)\] and \[\tan \left( \dfrac{\beta }{2} \right)\] are the roots of \[8{{x}^{2}}-26x+15=0\], where, \[a=8,b=-26\] and \[c=15\] then we have
\[\tan \left( \dfrac{\alpha }{2} \right)+\tan \left( \dfrac{\beta }{2} \right)=-\left( \dfrac{-26}{8} \right)=\dfrac{26}{8}\,\,\,......(1)\] and
\[\tan \left( \dfrac{\alpha }{2} \right)\times \tan \left( \dfrac{\beta }{2} \right)=\dfrac{15}{8}\,\,\,\,......(2)\]
We know that \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\],
Then, we have
\[\Rightarrow \tan \left( \dfrac{\alpha }{2}+\dfrac{\beta }{2} \right)=\dfrac{\tan \left( \dfrac{\alpha }{2} \right)+\tan \left( \dfrac{\beta }{2} \right)}{1-\tan \left( \dfrac{\alpha }{2} \right)\tan \left( \dfrac{\beta }{2} \right)}\]
Using equation (1) and (2) we have,
\[\Rightarrow \tan \left( \dfrac{\alpha }{2}+\dfrac{\beta }{2} \right)=\dfrac{\left( \dfrac{26}{8} \right)}{\left( 1-\left( \dfrac{15}{8} \right) \right)}\]
\[\Rightarrow \tan \left( \dfrac{\alpha }{2}+\dfrac{\beta }{2} \right)=\dfrac{\left( \dfrac{26}{8} \right)}{\left( \dfrac{8-15}{8} \right)}\]
\[\Rightarrow \tan \left( \dfrac{\alpha }{2}+\dfrac{\beta }{2} \right)=\dfrac{\left( \dfrac{26}{8} \right)}{\left( \dfrac{-7}{8} \right)}\]
Simplifying we have,
\[\Rightarrow \tan \left( \dfrac{\alpha +\beta }{2} \right)=-\dfrac{26}{7}\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right)\]
Now we also know that \[\cos 2A=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]
Then,
\[\Rightarrow \cos \left( \alpha +\beta \right)=\dfrac{1-{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}\]
Using equation (3) we know that,
\[\Rightarrow \cos \left( \alpha +\beta \right)=\dfrac{\left( 1-{{\left( -\dfrac{26}{7} \right)}^{2}} \right)}{\left( 1+{{\left( -\dfrac{26}{7} \right)}^{2}} \right)}\]
\[\Rightarrow \cos \left( \alpha +\beta \right)=\dfrac{\left( 1-\dfrac{676}{49} \right)}{\left( 1+\dfrac{676}{49} \right)}\]
Taking LCM and simplifying we have
\[\Rightarrow \cos \left( \alpha +\beta \right)=\dfrac{\left( \dfrac{49-676}{49} \right)}{\left( \dfrac{49+676}{49} \right)}\]
\[\Rightarrow \cos \left( \alpha +\beta \right)=\dfrac{-627}{725}\]
Or
\[\Rightarrow \cos \left( \alpha +\beta \right)=-\left( \dfrac{627}{725} \right)\].

Hence, the correct option is b) \[-\left( \dfrac{627}{725} \right)\]

Note: We must know the relation between the roots of the equation and the coefficients of the variable in a quadratic equation to solve this question. There are three half angle formulae of cosine:
$\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 = 1 - 2{\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}$
We can use all these formulae according to the conditions given in the question.