Question

In triangle ABC ,cos A +cos(B-C)=

Answer 1

Given :  triangle ABC

To Find : cos A +cos(B-C)

Solution:

In triangle ABC

A + B + C  = 180°

=> A = 180° - (B + C)

> CosA  = Cos(180° - (B + C))

=> CosA = - Cos(B + C)

cos A +cos(B-C)= - Cos(B + C)  + cos(B-C)

=> cos A +cos(B-C)=  cos(B-C) - Cos(B + C)  

=> cos A +cos(B-C)=  cos(B-C) - Cos(B + C)  

Using Cosx  - Cosy  = -2Sin{(x +y)/2}Sin{(x-y)/2}

=>  cos A +cos(B-C)= - 2Sin( B)Sin(-C)

Sin(-x) = - Sinx

=> cos A +cos(B-C)= - 2Sin( B)(-Sin(C)

=> cos A +cos(B-C)=  2Sin( B)Sin(C)

cos A +cos(B-C)=  2Sin( B)Sin(C)

Learn More:

If sin θ + cos θ = 2 , then evaluate : tan θ + cot θ - Brainly.in

https://brainly.in/question/7871635

sin θ cos(900-θ) + cos θ sin (900-θ)

https://brainly.in/question/10052334