Question

If $\sin A+\cos A=\sqrt{2}$ then $\sin A-\cos A$ is equal to:

Answer 1

$\sin A+\cos A=\sqrt{2}\left(\text{given}\right)$

Squaring both sides, we have

${(\sin A+\cos A)}^2={\left(\sqrt{2}\right)}^2$

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

$(1-{\cos }^{2}A)+(1-{\sin }^{2}A)+2\sin A\cos A=2(\text{Using identity}:{\sin }^2\theta +{\cos }^2\theta =1)$

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

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

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

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

$\Rightarrow {(\sin A-\cos A)}^2=0$

$\Rightarrow \sin A-\cos A=0$