Question

If $cos\theta +sin\theta =\sqrt{2}cos\theta$, then $cos\theta -sin\theta$=_____.

• A. $\sqrt{2}tan\theta$
• B. $\sqrt{2}sin\theta$
• C. $\frac{\sqrt{2}}{cos\theta +sin\theta }$
• D. $\sqrt{2}cos\theta$

Answer 1

Answer: (B)

$\sqrt{2}sin\theta$

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

or squaring on both sides gives,

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

$⟹{\cos }^2\theta -2\sin \theta \cos \theta ={\sin }^2\theta$

Now,$(cos\theta -\sin \theta {)}^2={\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta$

or from the above result, $(\cos \theta -sin\theta {)}^2={\sin }^2\theta +{\sin }^2\theta =2{\sin }^2\theta$

$or,\cos \theta -\sin \theta =\sqrt{2si{n}^2\theta }$

$or,\cos \theta -\sin \theta =\sqrt{2}\sin \theta$