Question

formula of sin(a+b).sin(a-b)

Answer 1

It will be

(sina+sinb)×(sina-sinb)

By using identity

${sina}^{2} - {sinb}^{2}$

hope it is helpful to you.

Answer 2

Answer:

$sin^2{a} - sin^2{b}$

Step by step explanation:

[since, sin (a+b) = sin a . cos b + cos a . sin b ,

           sin (a-b) = sin a . cos b - cos a . sin b .]

sin (a+b) . sin (a-b) = sin a . cos b + cos a . sin b . sin a . cos b - cos a . sin b

Above equation is in the form of '$a^2 - b^2$'

$a^2 - b^2 = (a+b) . (a-b)$

$= (sin a . cos b)^2 - (cos a . sin b)^2$

$= sin^2{a} . cos^2{b} - cos^2{a} . sin^2{b}$

$= sin^2{a} (1 - sin^2{b}) - (1 - sin^2{a}) sin^2{b}$

$= sin^2{a} - sin^2{a} . sin^2{b} - (sin^2{b} - sin^2{a} . sin^2{b})$

$= sin^2{a} - sin^{a} . sin^2{b} - sin^2{b} + sin^2{a} . sin^2{b}$

$[- sin^2{a} . sin^2{b}$ $, + sin^2{a} . sin^2{b}$ will be cancelled]

$= sin^2{a} - sin^2{b}$

           This is the answer........

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