A) \[-\frac{a}{b}\]
B) \[\frac{\sqrt{{{b}^{2}}-4ac}}{a}\]
C) \[1-\frac{a}{b}\]
D) \[1+\frac{{{a}^{2}}}{{{b}^{2}}}\]
E) \[\frac{a}{b}\]
Correct Answer: B
Solution :
Since,\[sec\theta \]and\[\tan \theta \]are the roots of the equation\[a{{x}^{2}}+bx+c=0\]. Then, \[\sec \theta +\tan \theta =-\frac{b}{a}\] and \[\sec \theta .\tan \theta =\frac{c}{a}\] Now, \[{{(\sec \theta -\tan \theta )}^{2}}\] \[={{(\sec \theta +\tan \theta )}^{2}}-4\sec \theta .\tan \theta \] \[={{\left( -\frac{b}{a} \right)}^{2}}-4\left( \frac{c}{a} \right)\] \[=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{4c}{a}\] \[\Rightarrow \] \[(\sec \theta -\tan \theta )=\frac{\sqrt{{{b}^{2}}-4ac}}{a}\]
You need to login to perform this action.
You will be redirected in
3 sec
