Question

Solve the equation: \[\left( 1-\tan \theta \right)\left( 1+\sin 2\theta \right)=1+\tan \theta \].

Answer 1

Hint: The formula for \[\sin 2\theta =\dfrac{2\tan \theta }{\left( 1+{{\tan }^{2}}\theta \right)}\]. Substitute it and simplify the expression until you get \[\left( 1+\tan \theta \right)\] and \[\left( -2{{\tan }^{2}}\theta \right)\] as equal to zero. From this find the value of \[\theta \] for both the cases and get the general equations.

Complete step-by-step answer:
We have been given which we need to solve.
\[\left( 1-\tan \theta \right)\left( 1+\sin 2\theta \right)=\left( 1+\tan \theta \right)\] - (1)
We know that \[\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }\]. Let us substitute this in (1).
\[\therefore \left( 1-\tan \theta \right)\left[ 1+\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right]=\left( 1+\tan \theta \right)\]
Now let us simplify the above expression,
\[\therefore \left( 1-\tan \theta \right)\left[ \dfrac{1+{{\tan }^{2}}\theta +2\tan \theta }{1+{{\tan }^{2}}\theta } \right]=\left( 1+\tan \theta \right)\] - (2)
We know that, \[{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\].
Looking in the above expression if x = 1 and \[y=\tan \theta \], we get,
\[{{\left( x+y \right)}^{2}}={{\left( 1+\tan \theta \right)}^{2}}=1+{{\tan }^{2}}\theta +2\tan \theta \]
Hence put \[{{\left( 1+\tan \theta \right)}^{2}}\] in place of \[\left( 1+{{\tan }^{2}}\theta +2\tan \theta \right)\] in equation 2. Thus we get,
\[\left( 1-\tan \theta \right)\dfrac{{{\left( 1+\tan \theta \right)}^{2}}}{1+{{\tan }^{2}}\theta }=1+\tan \theta \]
Let us apply cross multiplication property.
\[\therefore \left( 1-\tan \theta \right){{\left( 1+\tan \theta \right)}^{2}}=\left( 1+{{\tan }^{2}}\theta \right)\left( 1+\tan \theta \right)\] - (3)
We know that, \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
If a = 1 and b = \[\tan \theta \], then
\[\left( a-b \right)\left( a+b \right)=\left( 1-\tan \theta \right)\left( 1+\tan \theta \right)=\left( 1-{{\tan }^{2}}\theta \right)\]
Now, put \[\left( 1-\tan \theta \right)\left( 1+\tan \theta \right)=\left( 1-{{\tan }^{2}}\theta \right)\] in the LHS of (3).
\[\begin{align}
  & \left( 1+\tan \theta \right)\left( 1-{{\tan }^{2}}\theta \right)=\left( 1+{{\tan }^{2}}\theta \right)\left( 1+\tan \theta \right) \\
 & \Rightarrow \left[ \left( 1+\tan \theta \right)\left( 1-{{\tan }^{2}}\theta \right) \right]-\left[ \left( 1+{{\tan }^{2}}\theta \right)\left( 1+\tan \theta \right) \right]=0 \\
\end{align}\]
Let us take \[\left( 1+\tan \theta \right)\] as common from the above. We get,
\[\begin{align}
  & \left( 1+\tan \theta \right)\left[ \left( 1-{{\tan }^{2}}\theta \right)-\left( 1+{{\tan }^{2}}\theta \right) \right]=0 \\
 & \left( 1+\tan \theta \right)\left[ 1-{{\tan }^{2}}\theta -1-{{\tan }^{2}}\theta \right]=0 \\
 & \left( 1+\tan \theta \right)\left[ -2{{\tan }^{2}}\theta \right]=0 \\
\end{align}\]
Hence from the above we have,
\[1+\tan \theta =0\] or \[-2{{\tan }^{2}}\theta =0\]
In \[-2{{\tan }^{2}}\theta =0\]
\[\Rightarrow {{\tan }^{2}}\theta =0\]
Here \[{{\tan }^{2}}\theta \] is a positive quantity as it is a square and a positive quantity can only be zero if the term itself is zero.
\[\therefore \tan \theta =0\]
In \[1+\tan \theta =0\]
\[\Rightarrow \tan \theta =-1\]
When \[\tan \theta =0\] \[\Rightarrow \theta =0,\pi ,2\pi ,3\pi ,...\]
\[\tan \theta =-1\Rightarrow \theta =\dfrac{3\pi }{4},\dfrac{7\pi }{4},\dfrac{11\pi }{4},...,-\dfrac{\pi }{4}\]
Here \[\tan \theta \] can be zero only if the value of the \[\theta \] is \[0,\pi ,2\pi ,3\pi ,...\].
Similarly the value of \[\tan \theta \] can only be (-1) if the values of the \[\theta \] are \[\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{7\pi }{4},\dfrac{11\pi }{4},....\].
Thus taking generally, we have \[\tan \theta =\tan \alpha \].
Thus we can write the general equation as, \[\theta =n\pi +\alpha \].
Thus if we put, \[\alpha =-\dfrac{\pi }{4}\]
\[\theta =n\pi +\left( -\dfrac{\pi }{4} \right)\], where n = 0, 1, 2, …..
Or \[\theta =n\pi \], n = 0, 1, 2.
Thus we have the general solutions of the answers as,
\[\theta =n\pi -\dfrac{\pi }{4}\] and \[\theta =n\pi \]

Note: Here, \[\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}\].
If we put \[\tan x=\dfrac{\sin x}{\cos x}\] and simplify it, we will get the expression as \[\sin 2x=2\sin x\cos x\], which is an identity. Thus remember simple identities like this to make your solution easier.