વિધાન -1 : સમીકરણો x + left( {sin ,alph | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{\sin \alpha }&{\cos \alpha }\
1&{\cos \alpha }&{\sin \alpha }\
1&{ - \sin \

Vedclass · Accepted Answer

વિધાન $- 1$ સત્ય છે અને વિધાન $-2$ એ અસત્ય છે .

Vedclass · Answer

${\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{\sin \alpha }&{\cos \alpha }\
1&{\cos \alpha }&{\sin \alpha }\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} ight|$

$ = \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha  - \cos \alpha }&{\cos \alpha  - \sin \alpha }\
0&{\cos \alpha  + \sin \alpha }&{\sin \alpha  - \cos \alpha }\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} ight|$

$ = {\left( {\sin \alpha  - \cos \alpha } ight)^2} - \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } ight)$

$ = {\sin ^2}\alpha  - {\cos ^2}\alpha  - 2\sin \alpha .\cos \alpha  - {\cos ^2}\alpha  + {\sin ^2}\alpha $

$ = 2{\sin ^2}\alpha  - 2\sin \alpha .\cos \alpha $

$ = 2\sin \alpha \left( {\sin \alpha  - \cos \alpha } ight)$

Now, $\sin \alpha  - \cos \alpha  = 0$ for only

$\alpha  = \frac{\pi }{4}$ in $\left( {0,\frac{\pi }{2}} ight)$

${\Delta _1} = 2\left( {\sin \alpha } ight) 	imes 0 = 0$

Since value of $\sin \alpha $ is finite for $\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

Hence non- trivivial solution for only one value of $\alpha $ in $\left( {0,\frac{\pi }{2}} ight)$

$\left| {\begin{array}{*{20}{c}}
{\cos \alpha }&{\sin \alpha }&{\cos \alpha }\
{\sin \alpha }&{\cos \alpha }&{\sin \alpha }\
{\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} ight| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha }&{\cos \alpha }\
0&{\cos \alpha }&{\sin \alpha }\
{2\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} ight| = 0$

$ \Rightarrow 2\cos \alpha \left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } ight) = 0$

$	herefore \cos \alpha  = 0$ or ${\sin ^2}\alpha  - {\cos ^2}\alpha  = 0$

But $\cos \alpha  = 0$ not possible for any value of 

$\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

$	herefore {\sin ^2}\alpha  - {\cos ^2}\alpha  = 0 \Rightarrow \sin \alpha  =  - \cos \alpha $

which is also not possible for any value of 

$\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

Hence, there is no solution.

Similar Questions

સુરેખ સમીકરણ સંહતિ $a x+y+z=1$, $x+a y+z=1, x+y+a z=\beta$ માટે,નીચેના પૈકી કયું વિધાન સાચું નથી?

જો સુરેખ સમીકરણોની સંહતિ $x+y+2 z=6$, $2 x+3 y+a z=a+1$, $-x-3 y+b z=2 b$ જ્યાં $a, b \in R$, ને અસંખ્ય ઉકેલો હોય, તો $7 a+3 b=$ ______

સાબિત કરો કે બિંદુઓ $A(a, b+c), B(b, c+a), C(c, a+b)$ સમરેખ છે.

જો $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, તો $k =$