Question

If sin a, sin b and cos a are in GP, then roots of ${x^2} + 2x\cot b + 1 = 0$ are always
$
  A){\text{Equal}} \\
  B){\text{Real}} \\
  C){\text{Imaginary}} \\
  D){\text{Greater than 1}} \\
$

Answer 1

Hint: Here we will proceed by converting the sequence of geometric progression into quadratic equations. Then we will solve the quadratic equation to get the roots and check whether the roots are equal, real, imaginary or greater than 1. Therefore, we will get the desired result.

Complete step-by-step answer:
Geometric progression is a sequence of numbers where each term except the first is calculated by multiplying the previous one by a fixed, non-zero number called the common ratio.
Since sin a, sin b and cos a are in Geometric progression.
$\therefore {\sin ^2}b = \sin a\cos a \to (1)$
(Because we know that when three terms are in geometric progression,
Square of second term is product of first and third term.)
Also ${x^2} + 2x\cot b + 1 = 0$ (given)
We will solve the discriminant of this quadratic equation to find its roots.
Formula- $D = {b^2} - 4ac$
On comparison we get,
 a = 1
b = 2cotb
c = 1
Now , Substituting the values of a, b and c in formula , we get
$
  D = 4{\cot ^2}b - 4 \\
  D = 4\left[ {\dfrac{{{{\cos }^2}b}}{{{{\sin }^2}b}} - 1} \right] \\
  \because \left\{ {\dfrac{{\cos }}{{\sin }} = \cot } \right\} \\
   \Rightarrow D = 4\left[ {\dfrac{{{{\cos }^2}b - {{\sin }^2}b}}{{{{\sin }^2}b}}} \right] \\
   \Rightarrow D = 4\left[ {\dfrac{{1 - {{\sin }^2}b - {{\sin }^2}b}}{{{{\sin }^2}b}}} \right] \\
  \because \left\{ {{{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right\} \\
   \Rightarrow D = 4\left[ {\dfrac{{1 - 2{{\sin }^2}b}}{{{{\sin }^2}b}}} \right] \\
   \Rightarrow D = 4\left[ {\dfrac{{1 - 2\sin a\cos a}}{{{{\sin }^2}b}}} \right] \\
$
By using equation 1
$
   \Rightarrow D = 4\left[ {\dfrac{{{{\sin }^2}a + {{\cos }^2}a - 2\sin a\cos a}}{{{{\sin }^2}b}}} \right] \\
   \Rightarrow D = 4\left[ {\dfrac{{{{(\sin a - \cos a)}^2}}}{{{{\sin }^2}b}}} \right] \\
$
Since , ${(a - b)^2} = {a^2} + {b^2} - 2ab$
Above equation can also be written by taking all the terms inside the bracket as
$ \Rightarrow D = {\left[ {\dfrac{{2(\sin a - \cos a)}}{{\sin b}}} \right]^2}$
Now that we have the value of D , let us see the nature of roots on the basis of discriminant found. And we know,
(i) D > 0 so the roots of the quadratic equation are real and distinct.
 (ii) D = 0then the roots of the quadratic equation are real and equal.
(iii) D < 0 then the roots of the quadratic equation are unreal or imaginary.
Also we know that the square of any term (positive or negative) is always positive i.e. always greater than 0, Which means D > 0. Also $\sin b \ne 0$, otherwise the whole fraction will be undefined.
Thus, The roots of the given equation i.e. ${x^2} + 2x\cot b + 1 = 0$ are real.

So, the correct answer is “Option B”.

Note: Here in this question, one can face problems to solve the quadratic equation by quadratic formula so we can also use another way like completing the squares or factorization method. Also we must know the various formulas of trigonometry as we used it to simplify the quadratic equation.