Question

If $x = \sin \left( t \right)$ and $y = \tan \left( t \right)$, then $\dfrac{{dy}}{{dx}}$ is equal to
A. ${\cos ^3}\left( t \right)$
B. $\dfrac{1}{{{{\cos }^3}\left( t \right)}}$
C. $\dfrac{1}{{{{\cos }^2}\left( t \right)}}$
D. $\dfrac{1}{{{{\sin }^2}\left( t \right)}}$

Answer 1

Hint: To solve this question, we will first find the differentiation of $x$ and $y$ separately with respect to $t$. Then, in order to find $\dfrac{{dy}}{{dx}}$, we will divide both the equations obtained in the first step and then simplify the solution further to get the correct answer.

Complete step by step solution:
Given,
$x = \sin \left( t \right)$
Differentiating with respect to $t$, we get
$\dfrac{{dx}}{{dt}} = \cos \left( t \right)$ ……….$\left( 1 \right)$
And, $y = \tan \left( t \right)$
Differentiating with respect to $t$, we get
$\dfrac{{dy}}{{dt}} = {\sec ^2}\left( t \right)$ ……….$\left( 2 \right)$
Now, dividing equation $\left( 2 \right)$ by equation $\left( 1 \right)$ to find $\dfrac{{dy}}{{dx}}$,
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Substituting the values of $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$ from $\left( 2 \right)$ and $\left( 1 \right)$ respectively,
$\dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}\left( t \right)}}{{\cos \left( t \right)}}$
Simplifying the solution further, we get
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\cos }^2}\left( t \right) \times \cos \left( t \right)}}$ $\left[ {\because \sec \left( x \right) = \dfrac{1}{{\cos \left( x \right)}}} \right]$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\cos }^3}\left( t \right)}}$

Option ‘B’ is correct

Note: The key concept to solve this type of question is to first find out the differentiation of both $x$ and $y$ separately with respect to the trigonometric angle used in the question. Then divide and simplify it to get the correct answer. We should take care that the differentiation should be correct so as to be sure of the final answer.