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

In properties of triangles:-In ∆ABC 1+cos(A-B)cosC/1+cos(A-C)cosB = ?

In properties of triangles:-In ∆ABC 1+cos(A-B)cosC/1+cos(A-C)cosB = ?

Grade:11

1 Answers

Arun
25763 Points
3 years ago
Dear Rohan
 
2cos(A-B)cosC = cos(A-B+C) + cos(A-B-C) = cos(180-2B)+cos(2A-180) = - [cos2B+cos2A] 
In the second br B and C are interchanged so it is - [cos2C + cos2A] 
LHS = [1+cos(A-B) cosC] / [1+ cos(A-C) cosB] 
=[2+2cos(A-B) cosC] / [2+ 2cos(A-C) cosB] 
=[1 - cos2A +1 - cos2B]/[1 - cos2C +1 - cos2A] 
= [2sin^2A + 2sin^2B]/92sin^2C+2sin^2A]
(a^2 +b^2)/(a^2 + c^2) by using sine rule.
 
 
Regards
Arun (askIITians forum expert)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free