Question

If \[\tan \theta +\sin \theta =m\] and \[\tan \theta -\sin \theta =n\], prove that \[\left( {{m}^{2}}-{{n}^{2}} \right)=\pm 4\sqrt{mn}\].

Answer 1

Hint: First find the value of \[{{m}^{2}}\]and \[{{n}^{2}}\]individually. Then solve \[{{m}^{2}}-{{n}^{2}}\]and find its value. Similarly, find the value of \[\sqrt{mn}\]. Using the expression given in the question, prove that \[\left( {{m}^{2}}-{{n}^{2}} \right)\] and \[4\sqrt{mn}\] have similar expressions.

Complete step-by-step answer:
We have been given that, \[m=\tan \theta +\sin \theta \] and \[m=\tan \theta -\sin \theta \].
Now let us find the value of \[{{m}^{2}}\].
\[{{m}^{2}}={{\left( \tan \theta +\sin \theta \right)}^{2}}\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]. Similarly, find the value of \[{{\left( \tan \theta +\sin \theta \right)}^{2}}\].
\[\Rightarrow {{m}^{2}}={{\tan }^{2}}\theta +2\tan \theta \sin \theta +{{\sin }^{2}}\theta \]
Similarly, let us find the value of \[{{n}^{2}}\].
\[{{n}^{2}}={{\left( \tan \theta -\sin \theta \right)}^{2}}\]
We know that , \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]. Similarly, let us find the value of \[{{\left( \tan \theta -\sin \theta \right)}^{2}}\].
\[{{n}^{2}}={{\tan }^{2}}\theta -2\tan \theta \sin \theta +{{\sin }^{2}}\theta \]
Let us now, find the value of \[{{m}^{2}}-{{n}^{2}}\]. We know the expression for \[{{m}^{2}}\]and \[{{n}^{2}}\].
\[\begin{align}
  & {{m}^{2}}-{{n}^{2}}=\left( {{\tan }^{2}}\theta +2\tan \theta \sin \theta +{{\sin }^{2}}\theta \right)-\left( {{\tan }^{2}}\theta -2\tan \theta \sin \theta +{{\sin }^{2}}\theta \right) \\
 & {{m}^{2}}-{{n}^{2}}={{\tan }^{2}}\theta +2\tan \theta \sin \theta +{{\sin }^{2}}\theta -{{\tan }^{2}}\theta +2\tan \theta \sin \theta -{{\sin }^{2}}\theta \\
\end{align}\]
Let us cancel out the like terms.
\[\begin{align}
  & {{m}^{2}}-{{n}^{2}}=2\tan \theta \sin \theta +2\tan \theta \sin \theta \\
 & \therefore {{m}^{2}}-{{n}^{2}}=4\tan \theta \sin \theta -(1) \\
\end{align}\]
Now let us find the value of \[4\sqrt{mn}\].
\[4\sqrt{mn}=4\sqrt{\left( \tan \theta +\sin \theta \right)\left( \tan \theta -\sin \theta \right)}\]
We know that, \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
Similarly, \[4\sqrt{mn}=4\sqrt{{{\tan }^{2}}\theta -{{\sin }^{2}}\theta }\].
We know that, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\].
\[\therefore 4\sqrt{mn}=4\sqrt{\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta }\]
We know, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
\[\therefore {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
\[\begin{align}
  & \therefore 4\sqrt{mn}=4\sqrt{\dfrac{{{\sin }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }} \\
 & =4\sqrt{\dfrac{{{\sin }^{2}}\theta \left( 1-{{\cos }^{2}}\theta \right)}{{{\cos }^{2}}\theta }} \\
 & =4\sqrt{\dfrac{{{\sin }^{2}}\theta \times {{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}=4\sqrt{{{\sin }^{2}}\theta .\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} \\
 & =4\sqrt{{{\sin }^{2}}\theta .{{\tan }^{2}}\theta } \\
 & =\pm 4\sin \theta \tan \theta \\
 & \therefore 4\sqrt{mn}=\pm 4\tan \theta \sin \theta -(2) \\
\end{align}\]
By comparing the expression (1) and (2) we get that,
\[\begin{align}
  & {{m}^{2}}-{{n}^{2}}=4\tan \theta \sin \theta \\
 & 4\sqrt{mn}=4\tan \theta \sin \theta \\
\end{align}\]
\[\therefore \]We proved that, \[{{m}^{2}}-{{n}^{2}}=\pm 4\sqrt{mn}\].

Note: We have used basic trigonometric identities here to remember such identities which will enable us to solve questions like these easily. Don’t substitute the value of m and n directly in the expression given. Calculate LHS and RHS separately.