Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```        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.

```
8 years ago

Ramesh V
70 Points
```										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.
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,RameshIIT Kgp - 05 batch
```
8 years ago
mycroft holmes
266 Points
```										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

```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details