1Ans to Q1) z(xy - 1) = xy
Just to be sure, can someone give a solution to these:-
x=\sum_{n=0}^{\propto}{cos^{2n}\phi}
y=\sum_{n=0}^{\propto}{sin^{2n}\phi}
z=\sum_{n=0}^{\propto}{sin^{2n}\phi cos^{2n}\phi}
Then find the relation(s) between x, y and z.
Edit: 0<\phi <\frac{\pi }{2}
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7 Answers
1Q2)
S_{n}=\sum_{n=2}^{\propto}\frac{1}{n^{2}-1}
Find S_{n}
1S=\frac{1}{2}\sum_{r=2}^{\infty }{\frac{2}{r^{2}-1}}=\frac{1}{2}\sum_{r=2}^{\infty }{\frac{(1+r)-(r-1)}{r^{2}-1}}= \frac{1}{2}\sum_{r=2}^{\infty }{\frac{(1+r)-(r-1)}{(r+1)(r-1)}}=
=12(Σ∞r=2 (1r-1- 1r+1)
=12(1+12)=3/4
1708(1)
here x=\frac{1}{1-cos^{2}\theta }= \frac{1}{sin^{2}\theta }
sosin^{2}\theta =\frac{1}{x} .........(1)
similarly cos^{2}\theta =\frac{1}{y}.....(2)
and z=\frac{1}{1-sin^{2}\theta cos^{2}\theta }.....(3)
put value of sin^{2}\theta and cos^{2}\theta from equation (1) and (2) we get
xyz=xy+z.....(4)
but from adding equation (1) and (2)
we get sin^{2}\theta +cos^{2}\theta =1
\frac{1}{x}+\frac{1}{y}=1 or xy=x+y
replacing xy=x+y in equation ..(4) we get
xyz=x+y+z
1Thanx all!!!!
Answer to Q1) is given xyz= xy+z and xyz= x+yz
Answer to Q2) is given 4/3
BUt i'm pretty sure they are wrong..so thankuu.