Question

If x = a\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\} and $y = a\sin \theta$, then $\dfrac{{dy}}{{dx}}$ is equal to
A. $\cot \theta$
B. $\tan \theta$
C. $\sin \theta$
D. $\cos \theta$

Answer 1

First, we have to find the differentiation of both parameters $x$ and $y$ with respect to $\theta$. Then we will substitute the value of $\dfrac{{dy}}{{d\theta }}$ and $\dfrac{{dx}}{{d\theta }}$. Then we will divide both the equation to find the derivative $\dfrac{{dy}}{{dx}}$.

Complete step by step answer:

We are given that x = a\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\} and $y = a\sin \theta$.
First, we have to differentiate both parameters $x$ and $y$ with respect to $\theta$ to find the derivative $\dfrac{{dy}}{{dx}}$.

Now diving both the above differentiation, we get

$$\Rightarrow \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} \\\ \Rightarrow \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} \\\$$

Differentiating the parameter $x$ with respect to $\theta$, we get

$$\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \dfrac{{{{\sec }^2}\theta }}{{\tan \dfrac{\theta }{2}}} \times \dfrac{1}{2}} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \dfrac{{\cos \dfrac{\theta }{2}}}{{{{\cos }^2}\dfrac{\theta }{2}\sin \dfrac{\theta }{2}}} \times \dfrac{1}{2}} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \dfrac{1}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right) \\\$$

Using the property,$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ in the above equation, we get

$$\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \dfrac{1}{{\sin \theta }}} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( {\dfrac{{ - {{\sin }^2}\theta + 1}}{{\sin \theta }}} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( {\dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right){\text{ ......eq.(1)}} \\\$$

Differentiating the parameter $y$ with respect to $\theta$, we get

$\Rightarrow \dfrac{{dy}}{{d\theta }} = a\cos \theta {\text{ ......eq.(2)}}$

Dividing equation (2) by equation (1) to find $\dfrac{{dy}}{{dx}}$, we get

$$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\cos \theta \sin \theta }}{{a{{\cos }^2}\theta }} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin \theta }}{{\cos \theta }} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \tan \theta \\\$$

Hence, option B is correct.

Note: In this question, we have used basic trigonometric identities here to remember such identities, which will enable us to solve questions easily. Students should not substitute the value of $\dfrac{{dy}}{{d\theta }}$and $\dfrac{{dx}}{{d\theta }}$ directly to find the expression asked. Calculate $\dfrac{{dx}}{{d\theta }}$ and $\dfrac{{dy}}{{d\theta }}$ separately or else will get the wrong result. One should know the differentiation properties to solve this question.