Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

$</p><p>\left|\begin{array}{ccc}</p><p>1&cos A&sin A\\</p><p>1&cos B&sin B\\</p><p>1&cos C&sin C\end{array}</p><p>\right|=4sin\bigg(\frac{B-C}{2}\bigg)sin\bigg(\frac{C-A}{2}\bigg)sin\bigg(\frac{A-B}{2}\bigg)$

Apply

R1 -> R1-R3

R2 -> R2-R3

$\left|\begin{array}{ccc}</p><p>1-1&cos A-cos C&sin A- sin C\\</p><p>1-1&cos B-cos C&sin B-sin C\\</p><p>1&cos C&sin C\end{array}\right|$

$\left|\begin{array}{ccc}</p><p>0&cos A-cos C&sin A- sin C\\</p><p>0&cos B-cos C&sin B-sin C\\</p><p>1&cos C&sin C\end{array}\right|$

Expand the determinant along C1

= (cos A-cos C)(sin B-sin C)-(sin A-sin C)(cos B-cos C)

=cos A sin B-cos A sin C-cos C sin B+ cos C sin C-cos B sin A+cos C sin A+cos B sin C- cos C sin C

=cos A sin B-cos A sin C-cos C sin B-cos B sin A+cos C sin A+cos B sin C

=cos A sin B-cos B sin A+cos C sin A-cos A sin C+cos B sin C-cos C sin B

=sin(B-A)+sin(A-C)+sin(C-B)

if It is given that ABC is a triangle,than further it can be solved,just like you had solved in your previous class.

Answer 2

Step-by-step explanation:

$$\begin{lgathered}\left|\begin{array}{ccc} 1&cos A&sin A\\ 1&cos B&sin B\\ 1&cos C&sin C\end{array} \right|=4sin\bigg(\frac{B-C}{2}\bigg)sin\bigg(\frac{C-A}{2}\bigg)sin\bigg(\frac{A-B}{2}\bigg)\end{lgathered}$$

Apply

R1 -> R1-R3

R2 -> R2-R3

$$\begin{lgathered}\left|\begin{array}{ccc} 1-1&cos A-cos C&sin A- sin C\\ 1-1&cos B-cos C&sin B-sin C\\ 1&cos C&sin C\end{array}\right|\end{lgathered}$$

$$\begin{lgathered}\left|\begin{array}{ccc} 0&cos A-cos C&sin A- sin C\\ 0&cos B-cos C&sin B-sin C\\ 1&cos C&sin C\end{array}\right|\end{lgathered}$$

Expand the determinant along C1

= (cos A-cos C)(sin B-sin C)-(sin A-sin C)(cos B-cos C)

=cos A sin B-cos A sin C-cos C sin B+ cos C sin C-cos B sin A+cos C sin A+cos B sin C- cos C sin C

=cos A sin B-cos A sin C-cos C sin B-cos B sin A+cos C sin A+cos B sin C

=cos A sin B-cos B sin A+cos C sin A-cos A sin C+cos B sin C-cos C sin B

=sin(B-A)+sin(A-C)+sin(C-B)

if It is given that ABC is a triangle,than further it can be solved,just like you had solved in your previous class.