Question

4.

Answer 1

Answer:

See Explanation Below

Step-by-step explanation:

Given

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx.cos x.$

Required

Prove

To start with; we open the bracket on the left hand side

$(sin x - cos x)^2 = (sin x - cos x)(sin x - cos x)$

$(sin x - cos x)^2 = (sin x )(sin x - cos x)- (cos x)(sin x - cos x)$

$(sin x - cos x)^2 = sin^2 x -sinx cos x - sin xcos x + cos^2 x$

$(sin x - cos x)^2 = sin^2 x -2sinx cos x + cos^2 x$

Reorder

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

From trigonometry;

$sin^2x + cos^2x = 1$

So;

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

becomes

$(sin x - cos x)^2 = 1 - 2sinx cos x$

Also from trigonometry;

$sec^2x - tan^2x = 1$

So,  

$(sin x - cos x)^2 = 1 - 2sinx cos x$

becomes

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx cos x$

Proved