Question

sin A-cosA whole sqare = 1-sin2A
prove it​

Answer 1

Given,

$( {\sin(A) - \cos(A \: ) \: })^{2} = 1 - \sin(2A \: )$

Here,

L.H.S =

$( {\sin(A) - \cos(A \: ) \: })^{2}$

R.H.S =

$1 - \sin(2A \: )$

Now,

Solve the L.H.S :

$( {\sin(A) - \cos(A \: ) \: })^{2}$

$( {a - b})^{2} = {a}^{2} + {b}^{2} - 2ab$

$({ \sin(A \: ) })^{2} + ({ \cos(A \: ) })^{2} - 2 \sin(A \: ) \cos(A \: )$

${ \sin }^{2} A \: + { \cos }^{2} A \: - 2 \sin(A \: ) \cos(A \: ) \:$

$since \: { \sin }^{2} A \: + { \cos }^{2} A = 1$

$1 - 2 \: { \sin }A \: { \cos }A \:$

$since \: 2 \: { \sin }A \: { \cos }A \: \: = \sin \: 2A \:$

$1 - \sin \: 2A \:$

Therefore L.H.S = R.H.S.

Hence proved.