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,
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =3\sin \theta \cos \theta \\
 & \Rightarrow {{\cos }^{2}}\theta +2{{\sin }^{2}}\theta =3\sin \theta \cos \theta \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow 1+2{{\tan }^{2}}\theta =3\tan \theta \\
 & \Rightarrow 2{{\tan }^{2}}\theta -3\tan \theta +1=0 \\
\end{align}\]
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 \[\tan \theta \].