यदि y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}} \) तब \(\rm \frac{{dy}}{{dx}}\) का मान ज्ञात करें। 

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  1. \(\rm\frac{1}{2}\left( {secx.cotx + cosecx.tanx} \right)\)
  2. \(\rm \frac{1}{2}\left( {secx.cotx - cosecx.tanx} \right)\)
  3. \(\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)
  4. \(\rm \frac{1}{2}\left( {secx.tanx - cosecx.cotx} \right)\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)
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Detailed Solution

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उपयोग की गई अवधारणा:

त्रिकोणमिति सूत्र

sin 2x = 2sin x cos x

गणना:

y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}\)

y = \(\rm \frac{{\left( {sinx - cosx} \right)}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \frac{{sinx}}{{2.sinx.cosx}} - \;\frac{{cosx}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \left( {\frac{1}{{2cosx}}} \right)\; - \left( {\frac{1}{{2sinx}}} \right)\)

⇒ y = \(\frac{1}{2}\) (secx - cosecx)

दोनों पक्षों में अवकलन करने पर, हमें मिलता है

\(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]\)

\(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)

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