If \frac{\text{a²}}{\text{b+c}} =\frac{\text{b²}}{\text{c+a}} = \frac{\text{c²}}{\text{a+b}} = 1, then show that \frac{1}{\text{1+a}} + \frac{1}{\text{1+b}} + \frac{1}{\text{1+c}} = 1
WBBSE
Class X
Mathematics
Variation
03 Mark
2023
\frac{\text{a²}}{\text{b + c}} = \frac{\text{b²}}{\text{c + a}} = \frac{\text{c²}}{\text{a + b}} = 1
∴ a2 = b + c ; b2 = c + a ; c2 = a + b
\frac{1}{\text{1 + a}} + \frac{1}{\text{1 + b}} + \frac{1}{\text{1 + c}}= \frac{\text{a}}{\text{a + a²}} + \frac{\text{b}}{\text{b + b²}} + \frac{1}{\text{c + c²}}
= \frac{\text{a}}{\text{a + b + c}} + \frac{\text{b}}{\text{b + c + a}} + \frac{\text{c}}{\text{c + a + b}}
= \frac{\text{a + b + c}}{\text{a + b + c}} = 1 (Proved)
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