0

Guest

Courses New

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
25750 Points
6 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 ?

Other Related Questions on Trigonometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More