Question

Please help me with the question provided in the attachment..

No spams please.. /(^_^) ​

Answer 1

We know that,

$\cos(\alpha-\beta)=\cos\alpha\cdot \cos\beta+\sin\alpha\cdot \sin\beta$

Conversely,

$\sin\alpha\cdot\sin\beta+\cos\alpha\cdot \cos\beta=\cos(\alpha-\beta)$

We use this.

In addition, we use,

$1.\ \ \sin^2\theta+\cos^2\theta=1\\ \\ 2.\ \ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

Given that,

$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$

$\sin A+\sin B+\sin C=0\ \ \ \Longrightarrow\ \ \ (\sin A+\sin B+\sin C)^2=0\\ \\ \textsf{also,}\\ \\ \cos A+\cos B+\cos C=0\ \ \ \Longrightarrow\ \ \ (\cos A+\cos B+\cos C)^2=0$

Now,

$\small\begin{aligned}&(\sin A+\sin B+\sin C)^2+(\cos A+\cos B+\cos C)^2=0\\ \\ \Longrightarrow\ \ &\sin^2A+\sin^2B+\sin^2C+2(\sin A\cdot\sin B+\sin B\cdot\sin C+\sin C\cdot\sin A)+\\ &\cos^2A+\cos^2B+\cos^2C+2(\cos A\cdot\cos B+\cos B\cdot\cos C+\cos C\cdot\cos A)=0\\ \\ \Longrightarrow\ \ &\sin^2A+\cos^2A+\sin^2B+\cos^2B+\sin^2C+\cos^2C+2(\sin A\cdot\sin B+\\ &\cos A\cdot\cos B+\sin B\cdot\sin C+\cos B\cdot\cos C+\sin C\cdot\sin A+\cos C\cdot\cos A)=0\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &1+1+1+2(\cos(A-B)+\cos(B-C)+\cos(C-A))=0\\ \\ \Longrightarrow\ \ &3+2(\cos(B-C)+\cos(C-A)+\cos(A-B))=0\\ \\ \Longrightarrow\ \ &2(\cos(B-C)+\cos(C-A)+\cos(A-B))=-3\\ \\ \Longrightarrow\ \ &\cos(B-C)+\cos(C-A)+\cos(A-B)=\frac{-3}{2}\end{aligned}$

Hence   (d) -3/2   is the answer.

Answer 2

Option C

solution is attached above ....