Question:
Simplify: \(\dfrac{(\sin x – \cos x)^2}{1 – \sin x \cos x}\)
सरल कीजिए: \(\dfrac{(\sin x – \cos x)^2}{1 – \sin x \cos x}\)
Options:
- \(\dfrac{1 – 2\sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{1 + 2\sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{2 – \sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{3 – 2\sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{1 – 2\sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{1 + 2\sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{2 – \sin x \cos x}{1 – \sin x \cos x}\)
- \(\dfrac{3 – 2\sin x \cos x}{1 – \sin x \cos x}\)
Correct Answer: a)
Explanation:
Expand the numerator:
\((\sin x – \cos x)^2 = \sin^2 x + \cos^2 x – 2\sin x \cos x = 1 – 2\sin x \cos x\).
Thus, the expression becomes \(\dfrac{1 – 2\sin x \cos x}{1 – \sin x \cos x}\).
अंश का विस्तार करने पर:
\((\sin x – \cos x)^2 = \sin^2 x + \cos^2 x – 2\sin x \cos x = 1 – 2\sin x \cos x\).
अतः व्यंजक \(\dfrac{1 – 2\sin x \cos x}{1 – \sin x \cos x}\) बन जाता है।