badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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
Menu
Grade: 12

                        

if x and y are the solutions of 12 sinx + 5 cos x= 2 y^2-8y+21, the value of 3864 cot(xy/2) is?

6 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							LHS=12 sinx+5cosx</ sqrt(144+25)
=12sinx+5cosx</13
LHS</13
RHS=2(y^2-4y)+21
=2(y^2-4y+4)-8+21
=2(y-2)^2+13
hence RHS>/13
Roots of the equation exist if LHS=RHS=13
=> 2y^2-8y+21=13
=>2y^2-8y+8=0
=>y^2-4y+4=0
=>(y-2)^2=0
=>y=2
12sinx+5cosx=13
(12/13)sinx+(5/13)cosx=1
cos@sinx+sin@cosx=1 where @=arctan(5/12)
sin(x+@)=1=sin(pi/2)
x+arctan(5/12)=pi/2
x=pi/2-arctan(5/12)=arccot(5/12)
cotx=5/12
Hence 3864 cot(xy/2)=3864 cotx
=3864(5/12)=1610

Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details