Trigonometrical Equations 3 Question 10

11. For 0<φ<π/2, if x=n=0cos2nφ,y=n=0sin2nφ, z=n=0cos2nφsin2nφ, then

(1993, 2M)

(a) xyz=xz+y

(b) xyz=xy+z

(c) xyz=x+y+z

(d) xyz=yz+x

Integer Answer Type Questions

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Answer:

Correct Answer: 11. (b, c)

Solution:

  1. For 0<φ</π/2, we have

x=n=0cos2nφ=1+cos2φ+cos4φ+cos6φ+

It is clearly a GP with common ratio of cos2φ which is 1.

Hence, x=11cos2φ=1sin2φS=a1r,1<r<1

Similarly, y=1cos2φ

and

z=11sin2φcos2φ

Now, x+y=1sin2φ+1cos2φ

=cos2φ+sin2φcos2φsin2φ=1cos2φsin2φ

Again, 1z=1sin2φcos2φ=11xy

1z=xy1xyxy=xyzz

xy+z=xyz

Therefore, (b) is the answer from Eq. (i).

[putting the value of xy ]

xyz=x+y+z

Therefore, (c) is also the answer.



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