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Sin(A+B)Sin(A-B)

Prove that ...

Question

Prove that ${sin}^{2} A - {sin}^{2} B = sin (A + B) \times sin (A - B)$

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Solution

Expanding the R.H.S, we write,
$s i n (A + B) s i n (A - B) = (s i n A c o s B + c o s A s i n B) (s i n A c o s B - c o s A s i n B)$
= $s i n^{2} A c o s^{2} B - c o s^{2} A s i n^{2} B - s i n A c o s B c o s A s i n B + c o s A s i n B s i n A c o s B$
$s i n^{2} A c o s^{2} B - c o s^{2} A s i n^{2} B$
Substituting $c o s^{2} B = 1 - s i n^{2} B$ , $c o s^{2} A = 1 - s i n^{2} A$
$s i n^{2} A (1 - s i n^{2} B) - (1 - s i n^{2} A) s i n^{2} B$
= $s i n^{2} A - s i n^{2} A s i n^{2} B - s i n^{2} B + s i n^{2} A s i n^{2} B$
= $s i n^{2} A - s i n^{2} B$
Hence LHS=RHS

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Sin(A+B)Sin(A-B)

Standard XII Mathematics

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