Ответы
Ответ:
Hey there !
Solution :
We know the formulas for all the values in the above question. They are :
Sin ( A + B ) = Sin A . Cos B + Cos A . Sin B
Sin ( A - B ) = Sin A . Cos B - Cos A . Sin B
Given Equation :
Sin ( A + B ) * Sin ( A - B ) = Sin² A - Sin² B
Proof :
LHS :
Substituting the values from the formula we get,
Sin ( A + B ) * Sin ( A - B )
=> ( Sin A . Cos B + Cos A . Sin B ) * ( Sin A . Cos B - Cos A . Sin B )
=> Sin A . Cos B ( Sin A . Cos B - Cos A . Sin B ) +
Cos A . Sin B ( Sin A . Cos B - Cos A . Sin B )
=> ( Sin A . Cos B )² - ( Cos A . Sin B )²
=> ( Sin²A . Cos²B ) - ( Cos²A . Sin²B )
=> ( Sin²A ( 1 - Sin²B ) ) - ( ( 1 - Sin²A ) ( Sin²B )
=> Sin²A - Sin²A.Sin²B - ( Sin²B - Sin²A.Sin²B )
=> Sin²A - Sin²A.Sin²B - Sin²B + Sin²A.Sin²B
Sin²A.Sin²B gets cancelled. Then we get,
=> Sin²A - Sin²B
RHS = Sin²A - Sin²B
LHS = RHS
Hence Proved !
Hope my answer helped :-)