MCQ
જો $y = x.\tan \frac{x}{2}$ અને $A\frac{{dy}}{{dx}} - B = x$ તો $\frac{B}{A} = .........$
- A$\cot \frac{x}{2}$
- ✓$\tan \frac{x}{2}$
- C$\tan x$
- D$\cot x$
$y=x.tan\frac{x}{2}$
$\frac{dy}{dx}=x sec^2\frac{x}{2}\frac{1}{2}+tan\frac{x}{2}$
$\frac{dy}{dx}=\frac{x}{2cos^2\frac{x}{2}}+tan\frac{x}{2}$
$\frac{dy}{dx}=\frac{x}{1+cosx}+tan\frac{x}{2}$
$(1+\cos x)\frac{dy}{dx}=x+\tan\frac{x}{2}(1+\cos x)(\sin \& \cos)$ માં ફેરવો
$(1+cosx)\frac{dy}{dx}=x+sinx$
$(1+cosx)\frac{dy}{dx}-sinx=x$
$A=1+cosx$ અને $B=sinx$
$\frac{A}{B}=\frac{sinx}{1+cosx}=\frac{(2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}$
$\frac{A}{B}=tan\frac{x}{2}$
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(કે જ્યાં $\mathrm{C}$ એ સંકલન અચળાંક છે )