Click to Chat

1800-1023-196

+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
`        please find the least value  of cos^2-6sinacosa+2+3sin^2a `
2 months ago

Vikas TU
9468 Points
```							Dear student Cos^2Q - 6SinQ*CosQ + 3Sin^2Q + 2Cos^2Q + Sin^2Q +1 - Cos2Q -3Sin2Q => 1+1 - Cos2Q -3Sin2Q +2 => 4 - [Cos2Q-3sin2Q]Minimum value => 4 - sqrt((-1)^2 + (-3)^2 )=> 4- sqrt(10)Hope this helps
```
2 months ago
Arun
22984 Points
```							 maximum value is 3cos^2a-6sinacosa +2+3sin^2a=cos^2a +sin^2a +2sin^2a -6sina cosa=3 +2sin2a(sin2a-3)it will have maximum value when second term is maximum value at sin2a minimum is -1therefore 3+2*-1(-1-3)=11
```
2 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions