Question

if A= $\left|\begin{array}{ccc}1+cos\alpha & 1+sin\alpha & 1\\ 1+cos\beta & 1+sin\beta & 1\\ 1& 1& 1\end{array}\right|$
$\ne$ 0, then

• A. $\alpha =\beta$
• B. $\alpha \ne \beta +n\pi$, n being any integer
• C. $\alpha \ne \beta +\frac{\pi }{2}$
• D. $\alpha \ne \beta -\frac{\pi }{2}$

Answer 1

Answer: (B)

$\alpha \ne \beta +n\pi$, n being any integer

$A=\left|\begin{array}{ccc}1+\cos \alpha & 1+\sin \alpha & 1\\ 1+\cos \beta & 1+\sin \beta & 1\\ 1& 1& 1\end{array}\right|\ne 0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$A=\left|\begin{array}{ccc}\cos \alpha -\cos \beta & \sin \alpha -\sin \beta & 0\\ \cos \beta & \sin \beta & 0\\ 1& 1& 1\end{array}\right|\ne 0$
$(\cos \alpha -\cos \beta )\sin \beta -\cos \beta (\sin \alpha -\sin \beta )\ne 0$
go by option
if we put $\alpha =\beta ,\alpha =\beta +\frac{\pi }{2},\alpha =\beta -\frac{\pi }{2}$
$itbecomes0\alpha \ne \beta +n\pi isonlysoln.$