Question

if A+B=90° prove (cosA + cosB)²=1+2cosA.sinA​

Answer 1

Step-by-step explanation:

we have that,

A+B = 90°

B = 90°-A

proof--

LHS : ( cosA + cosB )^2

= [ cosA + cos(90-A) ]^2

= ( cosA + sinA )^2

= cos^2A + 2cosA.sinA +sin^2A

= sin^2A + cos^2A + 2cosA.sinA

= 1+2cosA.sinA

Therefore,LHS = RHS (proved)

Answer 2

Answer:

As given below

Step-by-step explanation:

Given that  

A + B = 90°  from this we can write  B = 90°- A

here we need to prove (cos A + cos B)²= 1 + 2 cos A.sin A​)

       take LHS part

  ⇒  [ cos A + cos B ]²    

  ⇒  [ cos A + cos (90°-A) ]²       [ apply B = 90°- A  ]  

  ⇒  [ cos A  + sin A ]²                [ cos (90-θ) = sin θ ]  

  ⇒  cos² A  + sin² A  + 2 cos A sin A     [ from (a+b)² = a² + b² + 2 ab ]  

  ⇒  1 + 2 cos A sin A                [ We know that cos² θ + sin² θ = 1 ]  

  ⇒ LHS = RHS

  ⇒ hence it is proved that (cos A + cos B)²= 1 + 2 cos A.sin A​)