Question

The product of all values of $x$ satisfying the equation {\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10\left| x \right| + 4}}{{{x^2} + 5\left| x \right| + 3}}} \right) = \cot \left\\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18\left| x \right|}}{{9\left| x \right|}}} \right)} \right\\} + \dfrac{\pi }{2} is
(A) $9$
(B) $- 9$
(C) $- 3$
(D) $- 1$

Answer 1

Hint : This type of equation will be solved by the use of the basic concept of inverse trigonometric and also angle property of trigonometric.

Complete step-by-step answer :
{\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10\left| x \right| + 4}}{{{x^2} + 5\left| x \right| + 3}}} \right) = \cot \left\\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18\left| x \right|}}{{9\left| x \right|}}} \right)} \right\\} + \dfrac{\pi }{2} . . . . (given)
We have to find the product of real value of $x$ which satisfies the equation.
So, the equation can be written as
\Rightarrow {\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}}} \right) = \cot \left\\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18x}}{{9x}}} \right)} \right\\} + \dfrac{\pi }{2} . . . . By $\left( {\left| x \right| = x} \right)$
By changing $\cos$ function to $\sin$ function in left hand side of the above equation we get,
\Rightarrow {\sin ^{ - 1}}\sin \left[ {\dfrac{\pi }{2} - \left( {\dfrac{{2{x^2} + 10x4}}{{{x^2} + 5x + 3}}} \right)} \right] = \cot \left\\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18x}}{{9x}}} \right)} \right\\} + \dfrac{\pi }{2} . . . . $\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)$
Applying the inverse trigonometric property in the both side we get,
$\dfrac{\pi }{2} - \left( {\dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}}} \right) = \dfrac{{2 - 18x}}{{9x}} + \dfrac{\pi }{2}$
On simplifying the above equation, we get,
$- \dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}} = \dfrac{{2 - 18x}}{{9x}}$
On cross multiplying the above equation we get,
$- 18{x^3} - 90{x^2} - 36x = 2{x^2} + 10x + 6 - 18{x^3} - 90x - 54x$
By rearranging above equation we get,
$\Rightarrow 2{x^2} - 8x + 6 = 0$
Now, factorize the above equation to find the value of $x$
$\Rightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0$
Since, there are two values of $x$
$x = 1$
$x = 3$
Therefore, the product of two real roots $= 3 \times 1 = 3$
Hence the product of the real roots which satisfies the given equation is $3$ .
Therefore from the above explanation the correct option is [] $3$ .

Note : In this question we use inverse trigonometric $\left[ {\cot .ca{t^{ - 1}}\left( \theta \right) = \theta \sin .{{\sin }^{ - 1}}\left( \theta \right)} \right]$
Also used $\cos \theta = \sin \left[ {\dfrac{\pi }{2} - \theta } \right]$ , we must be careful on this.