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 k sin2x + (1/k) cosec 2 x = 2, for all x belonging to (0, pie/2) then cos2x + 5 sinx cosx + 6 sin 2x is equals to
(a) (k2+5k+6) / k2
(b) (k2-5k+6) / k2
(c)6
(d) none of these```
8 years ago

Ramesh V
70 Points
```										as k sin2x + (1/k) cosec 2 x = 2
which is (ksin2x - 1)2 = 0
or sin2x = 1/k  , cos2x = 1 -(1/k)
for cos2x + 5 sinx cosx + 6 sin 2x
= 5 - 5/k +6/k + 5/k*(k-1)1/2
= ( 1+5k+ 5*(k-1)1/2) / k
ans: none of the these
--
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
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

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

Post Question