Question

What is z , $△ABC$
$\frac{\cos 2B-\cos 2C}{b+c}+\frac{\cos 2C-\cos 2A}{c+a}+\frac{\cos 2A-\cos 2B}{a+b}=z$

• A. 0
• B. 2
• C. 3
• D. 1

Answer 1

Answer: (A)

0

$\frac{\cos 2B-\cos 2C}{b+c}+\frac{\cos 2C-\cos 2A}{C+a}+\frac{\cos 2A-\cos 2B}{a+b}$

$=2$

Taking the this

$$\begin{array}{cc}\therefore \cos 2B& =2{\cos }^{2}B-1-\\ \cos 2c& =2{\cos }^{2}c-1\\ \cos 2A& =2{\cos }^{2}A-1\end{array}$$

and $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=t$ ( Let)

how, substituting the new value we get,

{ in LHS. }

$\Rightarrow \frac{2{\cos }^{2}B-1-(2{\cos }^{2}C-1)}{t(\sin B+\sin C)}+\frac{2{\cos }^{2}C-1-(2{\cos }^{2}A-1)}{t(\sin C+\sin A}+\frac{2{\cos }^{2}A-1-(2{\cos }^{2}B-1)}{t(\sin A+\sin B)}$

$\Rightarrow \frac{2}{t}[\frac{{\cos }^{2}B-{\cos }^{2}c}{\sin B+\sin c}+\frac{{\cos }^{2}c-{\cos }^{2}A}{\sin c+\sin A}+\frac{{\cos }^{2}A-{\cos }^2\beta }{\sin A+\sin B}$

$\begin{array}{cc}\because -{\cos }^{2}B& =1-{\sin }^{2}B\\ {\cos }^{2}C& =1-{\sin }^{2}C\\ {\cos }^{2}A& =1-{\sin }^{2}A\end{array}$

$\Rightarrow \frac{2}{t}[\frac{1-{\sin }^{2}B-(1-{\sin }^{2}c)}{\sin B+\sin c}+\frac{1-{\sin }^{2}c-[1-{\sin }^{2}A)}{\sin C+\sin A}$

$+\frac{1-{\sin }^{2}A-(1-{\sin }^{2}B)}{\sin A+\sin B}]$

$\begin{array}{cc}\Rightarrow & \frac{2}{t}[\sin c-\sin B+\sin A-\sin c+\sin B-\sin A\\ & =0.\therefore z=0\\ & \text{option}A=0\end{array}$