MCQ
$dy - \sin x\sin ydx = 0$ નો ઉકેલ મેળવો.
- ✓${e^{\cos x}}\tan \frac{y}{2} = c$
- B${e^{\cos x}}\tan y = c$
- C$\cos x\tan y = c$
- D$\cos x\sin y = c$
==> $\tan \frac{y}{2} = {e^{ - \cos x + c}}$ ==> ${e^{\cos x}}\tan \frac{y}{2} = {e^C} = c$.
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કારણ $(R) : \,$ જો $\overline {{\text{AB}}} \,\, = \,\,\vec a ,\;\,\overline {BC} \,\,\, = \,\,\vec b \,$ તો $\overline {AC} = \,\vec a + \,\,\vec b $ (સરવાળા ત્રિકોણ નિયમ )