Verified by Ajay Kumar
Simplify: \(\frac{1+\sin(x)}{1-\cos(x)}+\frac{1-\sin(x)}{1+\cos(x)}\)
\(\frac{2(5+\sin(x)\cos(x))}{4\sin^2(x)}\)
\(\frac{2(1+\sin(x)\cos(x))}{\sin^2(x)}\)
\(\frac{3(1+\sin(x)\cos(x))}{5\sin^2(x)}\)
\(\frac{6(3+\sin(x)\cos(x))}{\sin^2(x)}\)
Answer: B
Taking the LCM, the numerator becomes \((1+\sin x)(1+\cos x)+(1-\sin x)(1-\cos x)\), and the denominator becomes \((1-\cos x)(1+\cos x)\). Expanding the numerator gives \(2+2\sin x\cos x\). The denominator is \(1-\cos^2x=\sin^2x\). Therefore, the expression simplifies to \(\frac{2(1+\sin x\cos x)}{\sin^2x}\). Trick: Use the identity \(1-\cos^2x=\sin^2x\).
सरल करें: \(\frac{1+\sin(x)}{1-\cos(x)}+\frac{1-\sin(x)}{1+\cos(x)}\)
\(\frac{2(5+\sin(x)\cos(x))}{4\sin^2(x)}\)
\(\frac{2(1+\sin(x)\cos(x))}{\sin^2(x)}\)
\(\frac{3(1+\sin(x)\cos(x))}{5\sin^2(x)}\)
\(\frac{6(3+\sin(x)\cos(x))}{\sin^2(x)}\)
उत्तर: B
LCM लेने पर अंश \((1+\sin x)(1+\cos x)+(1-\sin x)(1-\cos x)\) और हर \((1-\cos x)(1+\cos x)\) होगा। अंश को खोलने पर \(2+2\sin x\cos x\) मिलता है। हर \(1-\cos^2x=\sin^2x\) होगा। अतः व्यंजक सरल होकर \(\frac{2(1+\sin x\cos x)}{\sin^2x}\) हो जाता है। ट्रिक: सूत्र \(1-\cos^2x=\sin^2x\) का प्रयोग करें।
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