Question

If $\sin \theta +\cos \theta =1$ then $(\sin \theta -\cos \theta )$ is equal to

• A. $\pm 1$
• B. $\frac{1}{2}$
• C. 2
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $\pm 1$

$sin\theta +cos\theta =1$
squaring both side, we get
$(sin\theta +cos\theta {)}^2=1$
$si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =1$
$si{n}^2\theta +co{s}^2\theta =1$
$\therefore 2sin\theta cos\theta =0$
$(sin\theta -cos\theta {)}^2=si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta$
$(sin\theta -cos\theta {)}^2=1$
$\therefore (sin\theta -cos\theta )=1,-1$