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: 12th pass
        
if tan a/2= square root (1+k/1-k)tan b/2 then sec a (cos b - k)/1-kcos b 
18 days ago

Answers : (1)

Arun
16143 Points
							
tana/2=√(1-k/(1+k) tanb/2
or, tanb/2=√(1+k)/(1-k) tana/2
squaring both sides, 
tan²b/2=(1+k)/(1-k) tan²a/2
or, (1-tan²b/2)/(1+tan²b/2)={(1-k)-(1+k)tan²a/2}/{(1-k)+(1+k)tan²a/2}
[by dividendo-componendo method]
or, cosb=(1-k-tan²a/2-ktan²a/2)/(1-k+tan²a/2+ktan²a/2)
or, cosb{(1-tan²a/2)-k(1+tan²a/2)}/{(1+tan²a/2)-k(1-tan²a/2)}
or, cosb=[{(1-tan²a/2)-k(1+tan²a/2)}/(1+tan²a/2)]/
                                                   [{(1+tan²a/2)-k(1-tan²a/2)}/(1+tan²a/2)]
or, cosb=(cosa-k)/(1-kcosa) [∵, (1-tan²a/2)/(1+tan²a/2)=cosa] 
 
 
hence on side transwpfer you will get amswer as 1
18 days ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


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


Get Extra Rs. 64 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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