Question

If $1+{{\sin }^{2}}\theta =3\sin \theta \cos \theta$, then prove that $\tan \theta =1$ or $\tan \theta =\dfrac{1}{2}$.

Answer 1

Hint: Put ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, in place of 1 in the given expression. Divide the entire expression by ${{\cos }^{2}}\theta$. Now you get a quadratic equation in terms of $\tan \theta$. Hence solve it with the help of a quadratic formula and find the value of $\tan \theta$.

Complete step-by-step answer:
We have been given a trigonometric function,
$\Rightarrow 1+{{\sin }^{2}}\theta =3\sin \theta \cos \theta$
We know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Hence replace 1 with (${{\sin }^{2}}\theta +{{\cos }^{2}}\theta$) in the above expression. Thus we get,

& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =3\sin \theta \cos \theta \\\ & \Rightarrow {{\cos }^{2}}\theta +2{{\sin }^{2}}\theta =3\sin \theta \cos \theta \\\ \end{aligned}$$ Now let us divide the entire expression by $${{\cos }^{2}}\theta $$. We get, $$\Rightarrow 1+\dfrac{2{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }=\dfrac{3\sin \theta }{\cos \theta }$$ We know that $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$$. Hence replace $$\dfrac{\sin \theta }{\cos \theta }$$ by $$\tan \theta $$. Thus we got the new expression as, $$\begin{aligned} & \Rightarrow 1+2{{\tan }^{2}}\theta =3\tan \theta \\\ & \Rightarrow 2{{\tan }^{2}}\theta -3\tan \theta +1=0 \\\ \end{aligned}$$ Let us put, x = $$\tan \theta $$. i.e. $$2{{x}^{2}}-3x+1=0$$ This expression is similar to the general quadratic equation $$a{{x}^{2}}+bx+c=0$$. Now by comparing both the equation, we get, a = 2, b = -3 and c = 1 Now let us put these values in the quadratic formula, $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ i.e. $$x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 2\times 1}}{2\times 2}=\dfrac{3\pm \sqrt{9-8}}{4}=\dfrac{3\pm 1}{4}$$ Hence, $$x=\dfrac{3+1}{4}=\dfrac{4}{4}=1$$ $$x=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}$$ Hence we got the values as x = 1 and x = $$\dfrac{1}{2}$$. We put x = $$\tan \theta $$. Thus we can say that, $$\tan \theta =1$$ and $$\tan \theta =\dfrac{1}{2}$$. Thus we proved that $$\tan \theta =1$$ or $$\tan \theta =\dfrac{1}{2}$$. Hence proved Note: You may put, $${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $$. Thus getting, $$2+{{\cos }^{2}}\theta =3\sin \theta \cos \theta $$. Thus you may complicate the solution by substituting in place of $${{\sin }^{2}}\theta $$. You should understand we need a quadratic equation with $$\tan \theta $$.