Question

If \[\sqrt{3}\tan \phi =3\sin \phi \], then find the value of \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi \].

Answer 1

Hint:First of all, consider the given equation and use \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Simplify this equation to get the value of \[\sin \phi \text{ and }\cos \phi \]. Now, from these values, get the corresponding \[\sin \phi \text{ and }\cos \phi \] and use these two values to get value of the expression \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi \].

Complete step-by-step answer:
Here, we are given that \[\sqrt{3}\tan \phi =3\sin \phi \]. We have to find the value of \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi \]. Let us consider the expression given in the question,
\[\sqrt{3}\tan \phi =3\sin \phi \]
We know that, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. By using this in the above equation, we get,
\[\sqrt{3}\dfrac{\sin \phi }{\cos \phi }=3\sin \phi \]
By transposing all the terms to the LHS of the above equation, we get,
\[\sqrt{3}\dfrac{\sin \phi }{\cos \phi }-3\sin \phi =0\]
By taking out \[\sqrt{3}\sin \phi \] common from the above equation, we get,
\[\sqrt{3}\sin \phi \left( \dfrac{1}{\cos \phi }-\sqrt{3} \right)=0\]
\[\Rightarrow \sqrt{3}\sin \phi \left( 1-\sqrt{3}\cos \phi \right)=0\]
From this, we get,
\[\sqrt{3}\sin \phi =0;\left( 1-\sqrt{3}\cos \phi \right)=0\]
\[\sin \phi =0;\sqrt{3}\cos \phi =1\]
\[\sin \phi =0;\cos \phi =\dfrac{1}{\sqrt{3}}\]
Case 1: Let us take \[\sin \phi =0\]
We know that,
\[{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1\]
So, we get,
\[0+{{\cos }^{2}}\phi =1\]
\[\cos \phi =\pm 1\]
Let us consider the expression asked in the question.
\[E={{\sin }^{2}}\phi -{{\cos }^{2}}\phi \]
By substituting \[\sin \phi =0\] and \[\cos \phi =\pm 1\], we get,
\[E={{\left( 0 \right)}^{2}}-{{\left( \pm 1 \right)}^{2}}\]
\[E={{\left( 0 \right)}^{2}}-{{\left( \pm 1 \right)}^{2}}\]
$E = - 1$
So, we get,
\[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =-1\]
Case 2: Let us take \[\cos \phi =\dfrac{1}{\sqrt{3}}\]. We know that,
\[{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1\]
So, we get,
\[{{\sin }^{2}}\phi +{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}=1\]
\[\Rightarrow {{\sin }^{2}}\phi +\dfrac{1}{3}=1\]
\[\Rightarrow {{\sin }^{2}}\phi =1-\dfrac{1}{3}\]
\[\Rightarrow {{\sin }^{2}}\phi =\dfrac{2}{3}\]
\[\sin \phi =\pm \sqrt{\dfrac{2}{3}}\]
Let us consider the expression asked in the question,
\[E={{\sin }^{2}}\phi -{{\cos }^{2}}\phi \]
By substituting \[\sin \phi =\pm \sqrt{\dfrac{2}{3}}\] and \[\cos \phi =\dfrac{1}{\sqrt{3}}\], we get,
\[E={{\left( \pm \sqrt{\dfrac{2}{3}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}\]
\[E=\dfrac{2}{3}-\dfrac{1}{3}\]
\[E=\dfrac{1}{3}\]
So, we get,
\[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =\dfrac{1}{3}\]

Note: In this question, students often make this mistake of considering just the first case where we get \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =-1\] and leave the second case where we get, \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =\dfrac{1}{3}\]. So this must be taken care of. Since, we have got two different values of \[\sin \phi \text{ and }\cos \phi \], two different values of \[{{\sin }^{2}}\phi -{{\cos }^{2}}\phi \] would also come.