Question

We are given a trigonometric expression, $2-{{\cos }^{2}}\theta =3\sin \theta \cos \theta ,\sin \theta \ne \cos \theta$ ,find the value of $\cot \theta$ .
a. $\dfrac{1}{2}$
b. $0$
c. $-1$
d. $2$

Answer 1

Hint: In the question we can see ${{\cos }^{2}}\theta$ and $\cot \theta$ which tells us that we can use trigonometric ratios and identity to solve the equation. Moreover, we have two terms containing $\cos \theta$ in which one is of second degree $\left( {{\cos }^{2}}\theta \right)$ other is of first degree \left\\{ 2\sin \theta \left( \cos \theta \right) \right\\}, so the quadratic equation can also be applicable.

Complete step-by-step answer:

We have been given the equation as $2-{{\cos }^{2}}\theta =3\sin \theta \cos \theta$ and we need to find the value of $\cot \theta$:
$\Rightarrow 2-{{\cos }^{2}}\theta =3\sin \theta \cos \theta$

Breaking $2$ into $1+1$ , we get:
$\Rightarrow 1+1-{{\cos }^{2}}\theta =3\sin \theta cos\theta$
$\Rightarrow 1+(1-{{\cos }^{2}}\theta )=3\sin \theta \cos \theta .........(i)$

Using the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we can rewrite the equation (i) as,
$({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+({{\sin }^{2}}\theta )=3\sin \theta \cos \theta$
$\Rightarrow 2{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =3\sin \theta \cos \theta$
$\Rightarrow 2{{\sin }^{2}}\theta -3\sin \theta \cos \theta +{{\cos }^{2}}\theta =0.............(ii)$
Now for the simplification purpose let’s put $\sin \theta =x$ and rewrite the equation (ii) as,
$\Rightarrow (2){{x}^{2}}-(3\cos \theta )x+({{\cos }^{2}}\theta )=0..........(iii)$

This equation is of quadratic form $a{{x}^{2}}+bx+c=0$ where $a=2,b=-3\cos \theta ,c={{\cos }^{2}}\theta$ .

Factoring the equation (iii), we get,
$2x(x-\cos \theta )-\cos \theta (x-\cos \theta )=0$
$\Rightarrow (2x-\cos \theta )(x-\cos \theta )=0.............(iv)$

But $x=\sin \theta$ , so the above equation (iv) becomes:
$(2\sin \theta -\cos \theta )(\sin \theta -\cos \theta )=0............(v)$

If the product of the two numbers is zero then any one number out of two numbers must be zero.

So, from the equation (v) we will have two solutions,
If $2\sin \theta -\cos \theta =0$
$\Rightarrow 2\sin \theta =\cos \theta$

On cross – multiplication, we get
$\Rightarrow \dfrac{\cos \theta }{\sin \theta }=2$

Now we know from trigonometric equation $\cot x=\dfrac{\sin x}{\cos x}$ , we get,
$\Rightarrow \cot \theta =2.......................(k)$
And if, $\sin \theta -\cos \theta =0$
$\Rightarrow \sin \theta =\cos \theta ................(l)$

But as per the condition specified in the question that $\sin \theta \ne \cos \theta$ so equation (l) becomes invalid and equation (k) is the only solution of the given problem.

Hence, the final answer is option (d).

Note: You may get stuck in the $2-{{\cos }^{2}}\theta =3\sin \theta \cos \theta$ solving for once if you will try to convert right hand side expression according to $\sin 2\theta =2\sin \theta \cos \theta$ formula because these terms seems something similar but solving this way will be very lengthy and time consuming.