Prove that $b^{2} cos 2 A - a^{2} cos 2 B = b^{2} - a^{2}$

Open in App

Solution

L.H.S $= b^{2} cos 2 A - a^{2} cos 2 B$
using trigonometric identities
$= b^{2} (1 - 2 {sin}^{2} A) - a^{2} (1 - 2 {sin}^{2} A)$
rearranging
$= (b^{2} - a^{2}) + 2 a^{2} {sin}^{2} B - 2 b^{2} {sin}^{2} A$
$= (b^{2} - a^{2}) + 2 a^{2} b^{2} / R^{2} - 2 b^{2} a^{2} / R^{2} = b^{2} - a^{2}$
$=$ R.H.S
Hence proved.

Suggest Corrections

Similar questions

△ A B C, b^{2} c o s 2 A - a^{2} c o s 2 B

P r o v e t h a t b^{2} cos 2 A - a^{2} cos 2 B = b^{2} - a^{2} .

△ A B C, b^{2} cos 2 A - a^{2} cos 2 B

b^{2} cos 2 A - a^{2} cos 2 A = b^{2} - a^{2}

a^{2} cos 2 B + b^{2} cos 2 A + 2 a b cos (A - B) = c^{2}

Join BYJU'S Learning Program