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

\[ \boxed{\pm\sqrt{2-a^2}} \]

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