\(V-I\) graph for conductor at temperature \(T_1\) and \(T_2\) is shown in the figure.


We know that resistance of a conductor is directly proportional to its temperature.
\(\begin{array}{ll}
\text {Hence, } & R_1 \propto T_1 \Rightarrow \tan \theta \propto T_1 \\
\Rightarrow & \tan \theta=K T_1 \quad \ldots (i) \\
\text {and } & R_2 \propto T_2 \\
\Rightarrow & \tan \left(90^{\circ}-\theta\right) \propto T_2 \\
\Rightarrow & \cot \theta \propto T_2 \\
\Rightarrow & \cot \theta=K T_2 \quad \ldots (ii)
\end{array}\)
\(\therefore\) From Eqs. (i) and (ii), we get
\(\begin{aligned}
& K T_2-K T_1=\cot \theta-\tan \theta \\
& K\left(T_2-T_1\right)=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \\
& \Rightarrow T_2-T_1=\frac{\cos ^2 \theta-\sin ^2 \theta}{K \sin \theta \cos \theta} \\
& \Rightarrow T_2-T_1=\frac{\cos 2 \theta}{\frac{1}{2} \sin 2 \theta} \Rightarrow T_2-T_1=\frac{2 \cot 2 \theta}{K} \\
& \Rightarrow T_2-T_1 \propto \cot 2 \theta \\
\end{aligned}\)