If $A =\begin{vmatrix}\sin (θ+α)&\cos (θ+α)&1\\\sin (θ+β)&\cos (θ+β)&1\\\sin (θ+γ)&\cos (θ+γ)&1\end{vmatrix}$, then
Answer & explanation
Correct answer: option 4
We have,
$A=\begin{vmatrix}\sin (θ+α)&\cos (θ+α)&1\\\sin (θ+β)&\cos (θ+β)&1\\\sin (θ+γ)&\cos(θ+γ)&1\end{vmatrix}$
$⇒A=\begin{vmatrix}\sin (θ+α)&\cos (θ+α)&1\\\sin (θ+β)-\sin (θ+α)&\cos (θ+β)-\cos (θ+α)&0\\\sin (θ+γ)-\sin (θ+α)&\cos(θ+γ)-\cos (θ+α)&0\end{vmatrix}$
Applying $R_2 → R_2 -R_1, R_3 → R_3 - R_1$
$⇒ A = \{\sin (θ+β)-\sin (θ+α)\} \{\cos(θ+γ)-\cos (θ+α)\} - \{\sin (θ+γ)-\sin (θ+α)\} \{\cos (θ+β)-\cos (θ+α)\}$
$⇒ A =\sin(\frac{β-α}{2})\cos(\frac{2θ+α+β}{2})\sin(\frac{2θ+α+γ}{2})\sin(\frac{α-γ}{2})-4\sin(\frac{2θ+α+γ}{2})\sin(\frac{2θ+α+γ}{2})\sin(\frac{α-β}{2})$
$⇒ A =4\sin(\frac{α-β}{2})\sin(\frac{γ-α}{2})\sin(\frac{γ-β}{2})$
Hence, A is independent of θ.