Question
\[ \text{If } \sin x+\cos x=a, \]
\[ \text{then } \sin x-\cos x=\ ? \]
Solution
Given,
\[ \sin x+\cos x=a \]
Squaring both sides,
\[ (\sin x+\cos x)^2=a^2 \]
\[ \sin^2x+\cos^2x+2\sin x\cos x=a^2 \]
\[ 1+2\sin x\cos x=a^2 \]
\[ 2\sin x\cos x=a^2-1 \]
Now,
\[ (\sin x-\cos x)^2 = \sin^2x+\cos^2x-2\sin x\cos x \]
\[ =1-(a^2-1) \]
\[ =2-a^2 \]
Therefore,
\[ \sin x-\cos x = \pm\sqrt{2-a^2} \]
Answer
\[ \boxed{\pm\sqrt{2-a^2}} \]