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  4. Assertion: | cos (θ+α) cos (θ+β) cos (θ+γ) sin (θ+α) sin (θ+β) sin (θ+γ) sin (β-γ) sin (γ-α) sin (α-β)| is independent of θ Reason: If f(θ)=c, then f(θ) is independent of θ.

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Q. Assertion: $\begin{vmatrix}\cos (\theta+\alpha) & \cos (\theta+\beta) & \cos (\theta+\gamma) \\ \sin (\theta+\alpha) & \sin (\theta+\beta) & \sin (\theta+\gamma) \\ \sin (\beta-\gamma) & \sin (\gamma-\alpha) & \sin (\alpha-\beta)\end{vmatrix} $ is independent of $\theta$
Reason: If $f(\theta)=c$, then $f(\theta)$ is independent of $\theta$.

Determinants

Solution:

Let $f(\theta)=\begin{vmatrix}\cos (\theta+\alpha) & \cos (\theta+\beta) & \cos (\theta+\gamma) \\ \sin (\theta+\alpha) & \sin (\theta+\beta) & \sin (\theta+\gamma) \\ \sin (\beta-\gamma) & \sin (\gamma-\alpha) & \sin (\alpha-\beta)\end{vmatrix}$
$\therefore f'(\theta)=\begin{vmatrix}-\sin (\theta+\alpha) & -\sin (\theta+\beta) & -\sin (\theta+\gamma) \\ \sin (\theta+\alpha) & \sin (\theta+\beta) & \sin (\theta+\gamma) \\ \sin (\beta-\gamma) & \sin (\gamma-\alpha) & \sin (\alpha-\beta)\end{vmatrix}$
$+\begin{vmatrix}\cos (\theta+\alpha) & \cos (\theta+\beta) & \cos (\theta+\gamma) \\ \cos (\theta+\alpha) & \cos (\theta+\beta) & \cos (\theta+\gamma) \\ \sin (\beta-\gamma) & \sin (\gamma-\alpha) & \sin (\alpha-\beta)\end{vmatrix}$
$+\begin{vmatrix}\cos (\theta+\alpha) & \cos (\theta+\beta) & \cos (\theta+\gamma) \\ \sin (\theta+\alpha) & \sin (\theta+\beta) & \sin (\theta+\gamma) \\ 0 & 0 & 0\end{vmatrix}$
$=0+0+0=0$
$ \Rightarrow f '(\theta)=0$
$ \Rightarrow f (\theta)= c$