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USE CODE: chait6

```				   If   β is such that sin2 β is not equals to zero, then show that the expression
[x2+2x cos2α+1]   / [x2+2x cosβ +1 ] for x belonging to R always lie between cos2α / cos2β  and sin2α/sin2β.

```

7 years ago

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```										Lets take y=[x2+2x cos2α+1]/[x2+2x cos2β +1] for x belonging to R
so, its [x2+2x cos2α+1] - y*[x2+2x cos2β +1] = 0
(1-y)x2+2x(cos2α-cos2β)+(1-y) = 0
the equation has real roots, when discriminant os positive
i.e., (cos2α-cos2β)2 - (1-y)2 > 0
[(cos2α-cos2β)+(1-y)].[(cos2α-cos2β)-(1-y)] > 0
[ y(1+cos2β)-(1+cos2α) ].[ y(1-cos2β)-(1-cos2α) ] < 0
(1+cos2α)/(1+cos2β)  <  y  <  (1-cos2α)/(1-cos2β)
so, y always lie between cos2α/cos2β < y < sin2α/sin2β.
--
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```
7 years ago
```										The given expression can be written as 1 + 2(cos 2A - cos 2B)/(x + 1/x + 2 cos 2B)

Since x + 1/x >= 2 or <= -2, we have  1/(cos 2B -1) <= 1/(x+1/x + 2 cos 2B) <= 1/2(1+cos 2B)

Hence 2(cos 2A - cos 2B)/(x + 1/x + 2 cos 2B) lies between 2(cos 2A - cos 2B)/2(1+cos 2B) and  2(cos 2A - cos 2B)/2(cos 2B-1)

so that 1 + 2(cos 2A - cos 2B)/(x + 1/x + 2 cos 2B) lies between 2(1+cos 2A)/2(1+cos 2B) and 2(1-cos 2A)/2(1-cos 2B) i.e.

between cos2A/cos2B and sin2A/sin2B

```
7 years ago

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