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).

& \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{aligned}$$ Let us take $$\left( 1+\tan \theta \right)$$ as common from the above. We get, $$\begin{aligned} & \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{aligned}$$ Hence from the above we have, $$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.