If a variable line in two adjacent positions has direction cosines l + δl, m + δm, n + δn then the small angle δθ between the two positions is given by:
(a) $\delta {{\theta }^{2}}=4\left( \delta {{l}^{2}}+\delta {{m}^{2}}+\delta {{n}^{2}} \right)$
(b) $\delta {{\theta }^{2}}=2\left( \delta {{l}^{2}}+\delta {{m}^{2}}+\delta {{n}^{2}} \right)$
(c) $\delta {{\theta }^{2}}=\left( \delta {{l}^{2}}+\delta {{m}^{2}}+\delta {{n}^{2}} \right)$
(d) None of these

Hint: First take the direction cosines of two adjacent lines which are (l, m, n) and (l + δl, m + δm, n + δn). Then use the formula ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$ for both the direction cosines then to find the small angle $\delta \theta $ use the formula for$\cos \delta \theta $ which is given as $\cos \delta \theta =l\left( l+\delta l \right)+m\left( m+\delta m \right)+n\left( n+\delta n \right)$.

Complete step-by-step answer:
Below are the direction cosines of two adjacent lines separated by $\delta \theta $:
(l, m, n) and (l + δl, m + δm, n + δn)
We know that sum of square of direction cosines is 1 so using this property in the above direction cosines we get,
${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$
${{\left( l+\delta l \right)}^{2}}+{{\left( m+\delta m \right)}^{2}}+{{\left( n+\delta n \right)}^{2}}=1$
Now, we are going to open the brackets of the above direction cosines and will get:
${{l}^{2}}+2l\delta l+{{\left( \delta l \right)}^{2}}+{{m}^{2}}+2m\delta m+{{\left( \delta m \right)}^{2}}+{{n}^{2}}+2n\delta n+{{\left( \delta n \right)}^{2}}=1$
Substituting ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$ in the above equation we get,
$\begin{align}

  & 1+2l\delta l+{{\left( \delta l \right)}^{2}}+2m\delta m+{{\left( \delta m

\right)}^{2}}+2n\delta n+{{\left( \delta n \right)}^{2}}=1 \\

 & \\

\end{align}$

1 will be cancelled out from both the sides we get,

$2\left( l\delta l+m\delta m+n\delta n \right)=-\left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right)$………….. Eq. (1)

Now, to find the small angle between adjacent lines we are using $\cos \delta \theta $ formula.

$\cos \delta \theta =l\left( l+\delta l \right)+m\left( m+\delta m \right)+n\left( n+\delta n \right)$

$\cos \delta \theta ={{l}^{2}}+{{m}^{2}}+{{n}^{2}}+l\delta l+m\delta m+n\delta n$

Substituting ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$ in the above equation we get,

$\cos \delta \theta =1+l\delta l+m\delta m+n\delta n$

Now, substituting the value of $l\delta l+m\delta m+n\delta n$ from eq. (1) in the above equation we get,

$\begin{align}

  & \cos \delta \theta =1-\dfrac{1}{2}\left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right) \\

 & \Rightarrow \left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right)=2\left( 1-\cos \delta \theta \right) \\

\end{align}$

Substituting the value of $\left( 1-\cos \delta \theta \right)=2{{\sin }^{2}}\dfrac{\delta \theta }{2}$in the above equation, we get:

$\left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right)=2.2\left( {{\sin }^{2}}\dfrac{\delta \theta }{2} \right)$

As it is given that δθ is a small angle so we can write the above equation as follows:

$\begin{align}
  & \left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right)=4{{\left( \dfrac{\delta \theta }{2} \right)}^{2}} \\

 & \Rightarrow \left( {{\left( \delta l \right)}^{2}}+{{\left( \delta m \right)}^{2}}+{{\left( \delta n \right)}^{2}} \right)={{\left( \delta \theta \right)}^{2}} \\

\end{align}$

From the above equation, we can say that:

Hence, the correct option is (c).

Note: For a very small angle θ, the sin of such a small angle θ is just θ.
$\sin \left( \theta \right)=\theta $. This relation we have used above in which we have written the expression as:${{\sin }^{2}}\left( \dfrac{\delta \theta }{2} \right)={{\left( \dfrac{\delta \theta }{2} \right)}^{2}}$