Question

If $\sin A+\cos A=\sqrt{2}$ then $\sin A-\cos A=$

Answer 1

Solve the required expression:

Given expression, $\sin A+\cos A=\sqrt{2}$…(i)

Squaring the above equation on both sides we get,

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

$\Rightarrow {\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A=2$ $[$applying formula ${\left(a+b\right)}^2=a^2+b^2+2ab$$]$

$\Rightarrow 1-{\cos }^{2}A+1-{\sin }^{2}A+2\sin A\cos A=2$ $[$$\because {\sin }^2\theta +{\cos }^2\theta =1$$]$

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

$\Rightarrow {(\sin A-\cos A)}^2=2-2$ $[$applying formula ${\left(a-b\right)}^2=a^2+b^2-2ab]$

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

Take the Square root of the above equation on both sides we get,

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

Hence, $\sin A-\cos A=0$.