Math, asked by ty009, 1 year ago

maths ch1 pg1.36 q18,19

plz explain in detail step by step ​

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ty009: ya sure
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ty009: what same???
ty009: oh ya I just saw
ty009: is it so? ok then don't solve . May be there is a problem in Q.

Answers

Answered by manish5365
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.
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ty009: dude I said both Q.s !!!
Answered by amitnrw
0

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


ty009: dear sir as two Q.s are given it is obvious that the user want ther answered to give ans. of both the Q.s so I request u to kindly resend both ans. by editing your current ans. or in whichever way u want.
ty009: *user want the answerer to give ans. of both
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