If $f(x)=\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec x & \tan x & x \\ 1 & \tan x-\tan \theta & 0\end{array}\right|$, then $f'(\theta)$ is

-1

We have,

$f(x)=\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec x & \tan x & x \\ 1 & \tan x-\tan \theta & 0\end{array}\right|$

$f'(x)=\left|\begin{array}{ccc}0 & 0 & 0 \\ \theta \sec x & \tan x & x \\ 1 & \tan x-\tan \theta & 0\end{array}\right| +\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec x \tan x & \sec ^2 x & 1 \\ 1 & \tan x-\tan \theta & 0\end{array}\right|+\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec x & \tan x & x \\ 0 & \sec ^2 x & 0\end{array}\right|$

∴  $f'(\theta)=0+\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec \theta \tan \theta & \sec ^2 \theta & 1 \\ 1 & 0 & 0\end{array}\right|+\left|\begin{array}{ccc}\sec \theta & \tan ^2 \theta & 1 \\ \theta \sec \theta & \tan \theta & \theta \\ 0 & \sec ^2 \theta & 0\end{array}\right|$

$\Rightarrow f'(\theta)=\tan ^2 \theta-\sec ^2 \theta+\sec ^2 \theta(\theta \sec \theta-\theta \sec \theta)=-1$