Question

If $\cos x = \left( {\dfrac{1}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$, and $\sin y = \left( {\dfrac{t}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$. Then what is the value of $\dfrac{{dy}}{{dx}}$?
A. 1
B. 0
C. $ - 1$
D. None of these

Answer 1

Hint: Substitute value of $t =\tan\theta $ and simplify both equations. Then differentiate both equations with respect to $t$. In the end, take the ratio of $\dfrac{{dy}}{{dt}}$ to $\dfrac{{dx}}{{dt}}$ and get the required answer.

Formula Used:
$\cos^{ - 1}\left( {\cos x} \right) = x$
$\sin^{ - 1}\left( {\sin x} \right) = x$
$\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$
$\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}$

Complete step by step solution:
 The given trigonometric equations are $\cos x = \left( {\dfrac{1}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$, and $\sin y = \left( {\dfrac{t}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$.
Let’s simplify both equations.
$\cos x = \left( {\dfrac{1}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$
$ \Rightarrow x = \cos^{ - 1}\left( {\dfrac{1}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$
Substitute $t = \tan\theta $ in the above equation.
$x = \cos^{ - 1}\left( {\dfrac{1}{{\sqrt {\left( {1 + \tan^{2}\theta } \right)} }}} \right)$
$ \Rightarrow x = \cos^{ - 1}\left( {\dfrac{1}{{\sqrt {\sec^{2}\theta } }}} \right)$ [Since $\sec^{2}\theta - 1 = \tan^{2}\theta $]
$ \Rightarrow x = \cos^{ - 1}\left( {\dfrac{1}{{\sec\theta }}} \right)$
$ \Rightarrow x = \cos^{ - 1}\left( {\cos \theta } \right)$ [Since $\dfrac{1}{{\sec x}} = \cos x$]
$ \Rightarrow x = \theta $ [Since $\cos^{ - 1}\left( {\cos x} \right) = x$]
Re-substitute the value of $\theta $ in the above equation.
$x = \tan^{ - 1}t$

Now differentiate the above equation with respect to $t$.
$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {\tan^{ - 1}t} \right)$
Apply the formula $\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
$\dfrac{{dx}}{{dt}} = \dfrac{1}{{1 + {t^2}}}$ $.....\left( 1 \right)$

Now simplify the given equation $\sin y = \left( {\dfrac{t}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$.
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{t}{{\sqrt {\left( {1 + {t^2}} \right)} }}} \right)$
Substitute $t = \tan\theta $ in the above equation.
$y = \sin^{ - 1}\left( {\dfrac{{\tan\theta }}{{\sqrt {\left( {1 + \tan^{2}\theta } \right)} }}} \right)$
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{\tan \theta }}{{\sqrt {\sec^{2}\theta } }}} \right)$ [Since $\sec^{2}\theta - 1 = \tan^{2}\theta $]
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{\tan \theta }}{{\sec \theta }}} \right)$
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{1}{{\cos \theta }}}}} \right)$ [Since $\dfrac{1}{{\sec x}} = \cos x$ and $\dfrac{{\sin x}}{{\cos x}} = \tan x$]
$ \Rightarrow y = \sin^{ - 1}\left( {\sin \theta } \right)$
$ \Rightarrow y = \theta $ [Since $\sin^{ - 1}\left( {\sin x} \right) = x$]
Re-substitute the value of $\theta $ in the above equation.
$y = \tan^{ - 1}t$

Now differentiate the above equation with respect to $t$.
$\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {\tan^{ - 1}t} \right)$
Apply the formula $\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
$\dfrac{{dy}}{{dt}} = \dfrac{1}{{1 + {t^2}}}$ $.....\left( 2 \right)$

Now apply the formula $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}$.
Substitute the equations $\left( 1 \right)$ and $\left( 2 \right)$ in the above formula.
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{1}{{1 + {t^2}}}}}{{\dfrac{1}{{1 + {t^2}}}}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = 1$

Option ‘A’ is correct

Note: Students often make the mistake of applying the formula to calculate $\dfrac{{dy}}{{dx}}$. Sometimes they used $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \cdot \dfrac{{dx}}{{dt}}$ which is an incorrect formula. The correct formula is $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$.