JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 15)
If an angle A of a $$\Delta $$ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :
secA, cotA
sinA, secA
secA, tanA
tanA, cosA
Explanation
Here, 9x2 + 27x + 20 = 0
$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$
$$ \Rightarrow $$$$\,\,\,$$ x = $${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$
$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$ $${4 \over 3}$$, $$-$$ $${5 \over 3}$$
Given, cosA = $$-$$ $${3 \over 5}$$
$$\therefore\,\,\,$$ sec A = $${1 \over {\cos A}}$$ = $$-$$ $${5 \over 3}$$
Here, A is an obtuse angle.
$$\therefore\,\,\,$$ tan A = $$-$$ $$\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$$
Hence, roots of the equation are sec A and tan A.
$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$
$$ \Rightarrow $$$$\,\,\,$$ x = $${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$
$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$ $${4 \over 3}$$, $$-$$ $${5 \over 3}$$
Given, cosA = $$-$$ $${3 \over 5}$$
$$\therefore\,\,\,$$ sec A = $${1 \over {\cos A}}$$ = $$-$$ $${5 \over 3}$$
Here, A is an obtuse angle.
$$\therefore\,\,\,$$ tan A = $$-$$ $$\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$$
Hence, roots of the equation are sec A and tan A.
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