Question

"Question32
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289
If A = B = 45° , verify that : sin (A - B) = sin A cos B - cos A sin B "

Answer 1

Hey there! Thanka for the question!

Given,
A = B = 45°

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

Substituting A = 45 , B = 45 °

sin( 45 - 45 ) = sin45 cos45 - cos45 sin45

sin0 = sin45 cos45 - cos45 sin45

We know, Trigonometry ratios of particular angles : sin0 = 0 , cos45 = 1/√2 , sin45 = 1/√2

0= 1/√2 ( 1/√2 ) + 1/√2 ( 1/√2)

0 = 1/2 - 1/2

0 = 0

Both Sides of the equation are equal.

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

Answer 2

Hey mate !!

Here's your answer !!

Given :

1. A = B = 45°

To verify :

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

Proof :

It is already given that A and B = 45°.

Hence we have to just substitute them in the equations given to verify them.

So after substituting we get,

Sin ( 45° - 45° ) = Sin 45° . Cos 45° - Cos 45° . Sin 45°

=> Sin ( 0° ) = Sin 45° . Cos 45° - Cos 45° . Sin 45°   -----( Eqn. 1 )

We know that,

Sin 0° = 0

Sin 45° = 1 / √ 2

Cos 45° = 1 / √ 2

Hence we must substitute the above values in Eqn. ( 1 )

So after substituting we get,

= 0 = 1 / √ 2 * 1 / √ 2 - 1 / √ 2 * 1 / √ 2

= 0 = 1 / 2 - 1 / 2

=> 0 = 0

=> LHS = RHS

Hence the given equations are verified.

Hope my answer helps you mate !!

Have a nice day ahead :-))