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  4. The angle made by the tangent of the curve x=a(t+ sin t cos t) ; y=a(1+ operatornamesint)2 with the x-axis at any point on it is

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Q. The angle made by the tangent of the curve $x=a(t+\sin t \cos t) ; y=a(1+\operatorname{sint})^2$ with the $x$-axis at any point on it is

Application of Derivatives

Solution:

$\frac{ dx }{ dt }= a +\frac{ a }{2} 2 \cos 2 t = a [1+\cos 2 t )=2 a \cos ^2 t$
$\frac{d y}{d t}=2 a(1+\sin t) \cdot \cos t $
$\frac{d y}{d x}=\frac{2 a(1+\sin t) \cdot \cos t}{2 a \cos ^2 t}=\frac{(1+\sin t)}{\cos t} $
$\tan \theta=\frac{(\cos (t / 2)+\sin (t / 2))^2}{\cos ^2(t / 2)-\sin ^2(t / 2)}=\frac{1+\tan \frac{t}{2}}{1-\tan \frac{t}{2}}=\tan \left(\frac{\pi}{4}+\frac{t}{2}\right) \Rightarrow \theta=\frac{\pi+2 t}{4}$