Question

The length of the subtangent at ‘$t$’ on the curve $x=a(t+\sin t)$, $y=a(1-\cos t)$ is

• A. $2a\sin \frac{t}{2}$
• B. $2a{\sin }^3\left(\frac{t}{2}\right)sec\left(\frac{t}{2}\right)$
• C. $a\sin t$
• D. $2a\sin \left(\frac{t}{2}\right)\tan \left(\frac{t}{2}\right)$

Answer 1

The correct option is C

$a\sin t$

Step 1 : Find the length of the tangent on the given curve

Given,

$x=a(t+\sin t)$

$\Rightarrow$ $\frac{dx}{dt}=a\left(1+\cos t\right)$

$y=a(1-\cos t)$

$\Rightarrow$ $\frac{dy}{dt}=a\sin t$

$\therefore$ $\frac{dy}{dx}=\frac{a\sin t}{a(1+\cos t)}$

$=\frac{2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{2{\cos }^2{\displaystyle \frac{t}{2}}-1+1}$ $$\begin{gathered} ...\left(\because sin2x=2sinxcosx\right) \\ \left(\because cos2x=2co{s}^{2}x-1\right) \end{gathered}$$

$\therefore$ $\frac{dy}{dx}=\tan \frac{t}{2}$

Step 2 : Find the length of the sub-tangent.

Length of the sub-tangent is given as $\frac{y}{y\text{'}}$

$\therefore$Sub-tangent $=\frac{y}{{\displaystyle \frac{dy}{dx}}}=\frac{a(1-\cos t)}{\tan {\displaystyle \frac{t}{2}}}$

$=\frac{a\left(2{\sin }^2{\displaystyle \frac{t}{2}}\right)\cos {\displaystyle \frac{t}{2}}}{\sin {\displaystyle \frac{t}{2}}}$ $...\left(\because cos2x=1-2si{n}^{2}x\right)$

$=2a\sin \frac{t}{2}\cos \frac{t}{2}$

$\therefore$ Sub-tangent $=a\sin t$

Hence option $\left(C\right)$ is the correct answer.