Question

If $\cos \theta -\sin \theta =1$, show that $\cos \theta +\sin \theta =1$

Answer 1

Given,

$\cos \theta -\sin \theta =1$

Squaring on both sides

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

${\cos }^2\theta +{\sin }^2\theta -2\cos \theta \sin \theta =1$

$1-2\cos \theta \sin \theta =1$

$2\cos \theta \sin \theta =0$

$(\cos \theta +\sin \theta {)}^2$

$={\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta$

$(\cos \theta +\sin \theta {)}^2=1+0$

$\therefore \cos \theta +\sin \theta =1$.