Question

If $x{\text{ }} = {\text{ }}2{\text{ }}log{\text{ }}cot{\text{ }}t$ and $y{\text{ }} = {\text{ }}tan{\text{ }}t{\text{ }} + {\text{ }}cot{\text{ }}t,$ $\dfrac{{dy}}{{dx}}\sin 2t + 1$ is equal to

{A){\text{ }}co{s^2}t} \\\ {B){\text{ }}si{n^2}t} \\\ {C){\text{ }}cos2t} \\\ {D){\text{ }}2co{s^2}t} \end{array}$$

Answer 1

Hint : In order to solve this question, we will start with differentiating the given equations with respect to t, then afterwards we will divide the two differential equations, now after substituting the values we will get the required answer.

Complete step-by-step answer :
Step 1: We have been given, $x{\text{ }} = {\text{ }}2{\text{ }}log{\text{ }}cot{\text{ }}t.$
On differentiating the above equation with respect to t, we get
$\dfrac{{dx}}{{dt}} = \dfrac{2}{{\cot t}}( - cose{c^2}t)$ $\ldots ..eq.\left( 1 \right)$
Also, we have been given, $y{\text{ }} = {\text{ }}tan{\text{ }}t{\text{ }} + {\text{ }}cot{\text{ }}t.$
On differentiating the above equation with respect to t, we get
$\dfrac{{dy}}{{dt}} = se{c^2}t - cose{c^2}t$ $\ldots ..eq.\left( 2 \right)$
Step 2: Now, dividing $eq.\left( 2 \right),$ by $eq.\left( 1 \right)$, we get
$\begin{gathered} \dfrac{{dy}}{{dx}} = \dfrac{{se{c^2}t{\text{ }}-{\text{ }}cose{c^2}t}}{{ - 2cose{c^2}t}}(\cot t) \\\ \dfrac{{dy}}{{dx}} = - \dfrac{{\cot t}}{2}({\tan ^2}t - 1) \\\ \dfrac{{dy}}{{dx}} = - \dfrac{1}{2}(\tan t - \cot t) \\\ \dfrac{{dy}}{{dx}} = - \dfrac{1}{2}(\dfrac{{\sin t}}{{\cos t}} - \dfrac{{\cos t}}{{\sin t}}) \\\ \dfrac{{dy}}{{dx}} = - \dfrac{1}{2}(\dfrac{{{{\sin }^2}t - co{s^2}t}}{{\sin t\cos t}}) \\\ \dfrac{{dy}}{{dx}} = \dfrac{{\cos 2t}}{{\sin 2t}} \\\ \end{gathered}$
Step: On substituting the value, $\dfrac{{dy}}{{dx}} = \dfrac{{\cos 2t}}{{\sin 2t}}$ in $\dfrac{{dy}}{{dx}}\sin 2t + 1$ , we get
$\dfrac{{dy}}{{dx}}\sin 2t + 1$ $=$ $\dfrac{{\cos 2t}}{{\sin 2t}}\sin 2t + 1$
$\begin{gathered} = \cos 2t + 1 \\\ = 2co{s^2}t \\\ \end{gathered}$
So, the correct answer is “Option D”.

Note : Before solving this question, students should learn/remind all the basic required formulas of trigonometry. More importantly one must be thorough with all the differentiation formulae.