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Answer 1

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Class 11

>>Applied Mathematics

>>Sequences and series

>>Geometric progression

>>For 0 < ϕ < pi/2 if x = ∑n ...

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For 0<ϕ<π/2 if x=∑

n=0

∞

cos

2n

ϕ,y=∑

n=0

∞

sin

2n

ϕ,z=∑

n=0

∞

cos

2n

ϕsin

2n

ϕ, then

Hard

Solution

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Correct option is

B

xyz=xy+z

Given x=

n=0

∑

∞

cos

2n

ϕ,y=

n=0

∑

∞

sin

2n

ϕ and z=

n=0

∑

∞

cos

2n

ϕsin

2n

ϕ

\since 0<ϕ<

2

π

, so each series is geometric series with common ratio r<1. Therefore, the series are convergent.

Now, x=

1−cos

2

ϕ

1

=

sin

2

ϕ

1

(∵S

∞

=

1−r

a

)

y=

1−sin

2

ϕ

1

(∵S

∞

=

1−r

a

)

=

cos

2

ϕ

1

z=

1−sin

2

ϕcos

2

ϕ

1

(∵S

∞

=

1−r

a

)

Consider, xyz=

sin

2

ϕcos

2

ϕ(1−sin

2

ϕcos

2

ϕ)

1

(1)

Also, =

sin

2

ϕcos

2

ϕ

1

+

1−sin

2

ϕcos

2

ϕ

1

xy+z=

sin

2

ϕcos

2

ϕ(1−sin

2

ϕcos

2

ϕ)

1−sin

2

ϕcos

2

ϕ+sin

2

ϕcos

2

ϕ

=

sin

2

ϕcos

2

ϕ(1−sin

2

ϕcos

2

ϕ)

1

=xyz [From(1)]