JEE Main On 16 April 2018 Question 8
Question: If an angle A of a $ \Delta ABC $ satisfies, then the roots of the quadratic $ 9x^{2}+27x+20=0 $ equation, are. [JEE Main 16-4-2018]
Options:
A) $ \sin A,\sec A $
B) $ \sec A,\tan A $
C) $ \tan A,\cos A $
D) $ secA,cotA $
Show Answer
Answer:
Correct Answer: B
Solution:
Using quadratic formula, the roots of the equation are,
$ 9x^{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.
