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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
$x \sin (a+y)+\sin a \cos (a+y)=0$ then show that $\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}$
✓

Answer

$x \sin (a+y)+\sin a \cos (a+y)=0$ $\ldots(i)$
$x \cdot \cos ( a +y) \cdot \frac{ d }{ d x}( a +y)+\sin ( a +y) \cdot \frac{ d }{ d x}(x)+\sin a [-\sin ( a +y)] \cdot \frac{ d }{ d x}( a +y)=0$
$\therefore x \cos ( a +y) \frac{ d y}{ d x}+\sin ( a +y)(1)-\sin ( a +y) \frac{ d y}{ d x}=0$
$\therefore[x \cos ( a +y)-\sin a \sin ( a +y)] \frac{ d y}{ d x}=-\sin ( a + y ) \quad \ldots \ldots . . \text { (ii) }$
From (i), we get
$x=\frac{-\sin a \cos ( a +y)}{\sin ( a +y)}$
Substituting the value of $x$ in (ii), we get
$\left[\frac{-\sin a \cos ( a +y)}{\sin ( a +y)} \cdot \cos ( a +y)-\sin a \sin ( a +y)\right] \frac{ d y}{ d x}=-\sin ( a + y )$
$\therefore-\sin a \left[\frac{\cos ^2( a +y)}{\sin ( a +y)}+\sin ( a +y)\right] \frac{ d y}{ d x}=-\sin ( a + y )$
$\therefore \frac{-\sin a \left[\cos ^2( a +y)+\sin ^2( a +y)\right]}{\sin ( a +y)} \frac{ d y}{ d x}-\sin ( a + y )$
$\therefore-\frac{\sin a (1)}{\sin ( a +y)} \cdot \frac{ d y}{ d x}=-\sin ( a + y )$
$\therefore \frac{ d y}{ d x}=\sin ( a +y)\left[\frac{\sin ( a +y)}{\sin a }\right]$
$\therefore \frac{ d y}{ d x}=\frac{\sin ^2( a +y)}{\sin a }$

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