A) \[\sin A,\sec A\]
B) \[\sec A,\tan A\]
C) \[\tan A,\cos A\]
D) \[\operatorname{secA},cotA\]
Correct Answer: B
Solution :
Using quadratic formula, the roots of the equation are, \[9{{x}^{2}}+27x+20=0\]are, \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] \[x=\frac{-27\pm \sqrt{{{27}^{2}}-4\times 9\times 20}}{2\times 9}\] \[x=-\frac{4}{3},-\frac{5}{3}\] Given, \[\cos A=-\frac{3}{5}.\] Hence,\[\sec A=\frac{1}{\cos A}=-\frac{5}{3}\] and \[\tan A=-\sqrt{{{\sec }^{2}}A-1}=-\frac{4}{3}\](since A is an obtuse angle, tan A will be negative). Thus, roots of the equation are \[\sec A\]and\[\tan A\]. Option B is correct.
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