Question

maths ch1 pg1.36 q18,19

plz explain in detail step by step ​

Answer 1

actually your answer of 18 is0 but it can't be solvedgo get the answer.

it's your answer of 19 in attachment dude hope it helps.

Answer 2

Answer:

$\frac{1}{1 + Z^{a-b} + Z^{a-c}} + \frac{1}{1 + Z^{b-c} + Z^{b-a}} + \frac{1}{1 + Z^{c-a} + Z^c-b}} = 1$

Step-by-step explanation:

Q 19

$\frac{1}{1 + Z^{a-b} + Z^{a-c}} + \frac{1}{1 + Z^{b-c} + Z^{b-a}} + \frac{1}{1 + Z^{c-a} + Z^c-b}}$

using

1 = zᵃ/Zᵃ  

Zᵃ⁻ᵇ = zᵃ/Zᵇ

Similarity others

= $\frac{1}{\frac{Z^a}{Z^a} + \frac{Z^a}{Z^b} + \frac{Z^a}{Z^c}} + \frac{1}{\frac{Z^b}{Z^b} + \frac{Z^b}{Z^c} + \frac{Z^b}{Z^a}} + \frac{1}{\frac{Z^c}{Z^c} + \frac{Z^c}{Z^a} + \frac{Z^c}{Z^b}}$

=$\frac{1}{Z^a}\times\frac{1}{\frac{1}{Z^a} + \frac{1}{Z^b} + \frac{1}{Z^c}} +\frac{1}{Z^b}\times\frac{1}{\frac{1}{Z^a} + \frac{1}{Z^b} + \frac{1}{Z^c}} + \frac{1}{Z^c}\times\frac{1}{\frac{1}{Z^a} + \frac{1}{Z^b} + \frac{1}{Z^c}}$

$=(\frac{1}{Z^a} + \frac{1}{Z^b} + \frac{1}{Z^c}} )\times(\frac{1}{\frac{1}{Z^a} + \frac{1}{Z^b} + \frac{1}{Z^c}}) \\ \\= 1$