Correct Answer - Option 3 :
\(\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)
Concept used:
Trigonometry formula
sin 2x = 2sin x cos x
Calculation:
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)
Differentiate both sides, we get
⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]\)
⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)
