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

Hint: 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.