MCQ
$\sin ^{-1}\left[\frac{\sin x-\cos x}{\sqrt{2}}\right]=\ldots \ldots \ldots\left(\right.$ જ્યાં $\left.-\frac{\pi}{4} \right)$
- A$x+\frac{\pi}{4}$
- ✓$x-\frac{\pi}{4}$
- C$\frac{1}{\sqrt{2}}$
- Dx
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$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ નો વિસ્તાર મેળવો.