Question

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

• A. $\frac{1}{4}(\pi +2t)$
• B. $\frac{1-\sin t}{\cos t}$
• C. $\frac{1}{4}(2t-\pi )$
• D. $\frac{1+\sin t}{\cos 2t}$

Answer 1

The correct option is A $\frac{1}{4}(\pi +2t)$

$x=a(t+sintcost)$

$\frac{dx}{dt}=2aco{s}^{2}t$

$y=a(1+sint{)}^2$

$\frac{dy}{dt}=2a(cost+costsint)$

$\Rightarrow \frac{dy}{dx}=\frac{1+sint}{cost}$

$=\frac{(cos\frac{t}{2}+sin\frac{t}{2}{)}^2}{co{s}^2\frac{t}{2}-si{n}^2\frac{t}{2}}$

$=\frac{1+tan\frac{t}{2}}{1-tan\frac{t}{2}}$

$=tan(\frac{\pi }{4}+\frac{t}{2})$

Slope of tangent to the curve $=tan(\frac{\pi }{4}+\frac{t}{2})$

So, the angle made by the tangent to the curve with the x-axis $=\frac{\pi }{4}+\frac{t}{2}$ or $\frac{1}{4}(\pi +2t)$