If $m s i n^{2} A = c o s^{2} (A - B) + c o s^{2} B - 2 c o s (A - B) c o s A c o s B .$ Find $m$ .

Open in App

Solution

$c o s^{2} (A - B) + c o s^{2} B - 2 c o s (A - B) c o s A c o s B$
$= c o s^{2} B + c o s^{2} (A - B) - 2 c o s (A - B) c o s A c o s B$
$= c o s^{2} B + c o s (A - B) {c o s (A - B) - 2 c o s A c o s B}$
$= c o s^{2} B + c o s^{2} B {c o s A c o s B + s i n A s i n B - 2 c o s A c o s B}$
$= c o s^{2} B + c o s (A - B) {s i n A s i n B - c o s A c o s B}$
$= c o s^{2} B - c o s (A - B) {c o s A c o s B - s i n A s i n B}$
$= c o s^{2} B - c o s (A - B) c o s (A + B)$
$= c o s^{2} B - (c o s^{2} A - s i n^{2} B)$
$= c o s^{2} B + s i n^{2} B - c o s^{2} A$
$= 1 - c o s^{2} A$
$= s i n^{2} A$
$\Rightarrow m = 1$

Suggest Corrections

Similar questions

s i n^{2} A = c o s^{2} (A + B) + c o s^{2} B - 2 c o s (A - B) c o s A c o s B

{sin}^{2} A = {cos}^{2} (A - B) + {cos}^{2} B - 2 cos (A - B) cos A cos B

{cos}^{2} (A - B) + {cos}^{2} B - 2 cos (A - B) cos A cos B

c o s^{2} (A - B) + c o s^{2} B - 2 c o s (A - B) c o s A c o s B

Δ

{cos}^{2} \frac{A}{2} + {cos}^{2} \frac{B}{2} - {cos}^{2} \frac{C}{2} = 2 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}

Join BYJU'S Learning Program