Question

"Question34
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289
If A = B = 45° ,verify that :cos (A - B) = cos A cos B + sin A sin B "

Answer 1

Hey there! Thanka for the question!

Given,
A = B = 45°

Now,
cos(A - B) = cosA cosB + sinA sinB

Substituting A = 45 , B = 45 °

cos( 45 - 45 ) = cos45 cos45 + sin45 sin45

cos0 = cos²45 + sin²45

We know, Trigonometry ratios of particular angles : cos0 = 1 , cos45 = 1/√2 , sin45 = 1/√2

1= (1/√2)² + (1/√2)²

1 = 1/2 + 1/2

1 = 1

Both Sides of the equation are equal.

Hence, We proved and verified that cos (A - B) = cos A cos B + sin A sin B holds good for A = B = 45°

Answer 2

Hey mate !!

Here's your answer !!

Given:

A = B = 45°

To verify:

Cos ( A - B ) = Cos A . Cos B + Sin A . Sin B

Proof :

Now we know the values of A and B. Hence substitute them in the equation.

= Cos ( 45 - 45 ) = Cos 45 . Cos 45 + Sin 45 > Sin 45

= Cos 0 = Cos² 45 + Sin² 45   -----( Equation 1 )

We know that,

Cos 0 = 1

Cos 45 = 1 / √ 2

Sin 45 = 1 / √ 2

Substitute them in Equation 1. After substituting we get,

= 1 = ( 1 / √ 2 )² + ( 1 / √ 2 )²

=> 1 = 1 / 2 + 1 / 2

=> 1 = 1

Hence LHS = RHS.

Hence verified.

Hope it helped you mate !!

Cheers !!