Trigonometric Functions — MATHS STD 11 Science — Question

4. $0$

Solution:

Using sine rule, we have

$\sum\text{a}^2(\sin\text{B}-\sin\text{C})$

$\text{a}^2\Big(\frac{\text{b}}{\text{k}}-\frac{\text{c}}{\text{k}}\Big)+\text{b}^2\Big(\frac{\text{c}}{\text{k}}-\frac{\text{a}}{\text{k}}\Big)+\text{c}^2\Big(\frac{\text{a}}{\text{k}}-\frac{\text{b}}{\text{k}}\Big)$

$=\frac{1}{\text{k}}(\text{a}^2\text{b}-\text{a}^2\text{c}+\text{b}^2\text{c}-\text{b}^2\text{a}+\text{c}^2\text{a}-\text{c}^2\text{b})$

This expression cannot be simplified to match with any of the given options.

However, if the quesion is "In any $\triangle\text{ABC},\sum\text{a}^2(\sin\text{B}-\sin\text{C})=",$ then the solution is as follows.

Using sine rule, we have

$\sum\text{a}^2(\sin^2\text{B}-\sin^2\text{C})$

$\text{a}^2\Big(\frac{\text{b}^2}{\text{k}^2}-\frac{\text{c}^2}{\text{k}^2}\Big)+\text{b}^2\Big(\frac{\text{c}^2}{\text{k}^2}-\frac{\text{a}^2}{\text{k}^2}\Big)+\text{c}^2\Big(\frac{\text{a}^2}{\text{k}^2}-\frac{\text{b}^2}{\text{k}^2}\Big)$

$\frac{1}{\text{k}^2}(\text{a}^2\text{b}^2-\text{a}^2\text{c}^2+\text{b}^2\text{c}^2-\text{b}^2\text{a}^2+\text{c}^2\text{a}^2-\text{c}^2\text{b}^2)$

$=\frac{1}{\text{k}^2}\times0$

$=0$

Hence, the correct answer is option (d).