Math, asked by jjmarzook4460, 11 months ago

Without expanding, prove that
| cosα sinα cos(α+δ) |
| cosβ sinβ cos(β+δ)| =0
| cosγ sinγ cos(γ+δ) |

Answers

Answered by MaheswariS
0

Answer:

\bf\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos(\alpha+\delta)\\cos\beta&sin\beta&cos(\beta+\delta)\\cos\gamma&sin\gamma&cos(\gamma+\delta)\end{array}\right|=0

Step-by-step explanation:

\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos(\alpha+\delta)\\cos\beta&sin\beta&cos(\beta+\delta)\\cos\gamma&sin\gamma&cos(\gamma+\delta)\end{array}\right|

using

\boxed{cos(A+B)=cosA\:cosB-sinA\:sinB}

=\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos\alpha\:cos\delta-sin\alpha\:sin\delta\\cos\beta&sin\beta&cos\beta\:cos\delta-sin\beta\:sin\delta\\cos\gamma&sin\gamma&cos\gamma\:cos\delta-sin\gamma\:sin\delta\end{array}\right|

By using properties of determinants

=\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos\alpha\:cos\delta\\cos\beta&sin\beta&cos\beta\:cos\delta\\cos\gamma&sin\gamma&cos\gamma\:cos\delta\end{array}\right|-\left|\begin{array}{ccc}cos\alpha&sin\alpha&sin\alpha\:sin\delta\\cos\beta&sin\beta&sin\beta\:sin\delta\\cos\gamma&sin\gamma&sin\gamma\:sin\delta\end{array}\right|

\text{Taking }cos\delta\text{ common from }C_3\text{ in first determinant}

=cos\delta\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos\alpha\\cos\beta&sin\beta&cos\beta\\cos\gamma&sin\gamma&cos\gamma\end{array}\right|-\left|\begin{array}{ccc}cos\alpha&sin\alpha&sin\alpha\:sin\delta)\\cos\beta&sin\beta&sin\beta\:sin\delta\\cos\gamma&sin\gamma&sin\gamma\:sin\delta\end{array}\right|

\text{Taking }sin\delta\text{ common from }C_3\text{ in second determinant}

=cos\delta\:\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos\alpha\\cos\beta&sin\beta&cos\beta\\cos\gamma&sin\gamma&cos\gamma\end{array}\right|-sin\delta\;\left|\begin{array}{ccc}cos\alpha&sin\alpha&sin\alpha\\cos\beta&sin\beta&sin\beta\\cos\gamma&sin\gamma&sin\gamma\end{array}\right|

using

\boxed{\text{If any two rows or columns of a determinat are equal, then the value of the determinant is zero}}

=cos\delta(0)-sin\delta(0)

=0

\implies\:\bf\left|\begin{array}{ccc}cos\alpha&sin\alpha&cos(\alpha+\delta)\\cos\beta&sin\beta&cos(\beta+\delta)\\cos\gamma&sin\gamma&cos(\gamma+\delta)\end{array}\right|=0

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