Question

If \[\tan y = 2t/\left( {1 - {t^2}} \right)\] and \[\sin x = 2t/\left( {1 + {t^2}} \right)\] then \[\dfrac{{dy}}{{dx}} = \]
(a) \[2/\left( {1 - {t^2}} \right)\]
(b) \[1/\left( {1 + {t^2}} \right)\]
(c) 1
(d) 2

Answer 1

Hint: To solve for\[\dfrac{{dy}}{{dx}}\], substitute \[t = \tan \theta \] in the given equations. Simplify the equation and use the formula \[\tan 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 - {{\tan }^2}\theta } \right)}}\] to differentiate y with respect to t. Use the formula \[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\] to simplify and differentiate x with respect to t. Divide both the derivatives i.e. \[\dfrac{{dy/dt}}{{dx/dt}}\] to get\[\dfrac{{dy}}{{dx}}\].

Formula Used:
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 - {{\tan }^2}\theta } \right)}}\]
\[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]

Complete step by step Solution:
Formulas to keep in mind before solving the problem:\[\tan 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 - {{\tan }^2}\theta } \right)}}\] and \[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\].
Given: \[\tan y = \dfrac{{2t}}{{\left( {1 - {t^2}} \right)}}\] -- equation (1)
Put \[t = \tan \theta \] in equation (1)
Therefore, equation (1) becomes
\[\tan y = \dfrac{{2\tan \theta }}{{\left( {1 - {{\tan }^2}\theta } \right)}}\]---equation (2)
We know that \[\tan 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 - {{\tan }^2}\theta } \right)}}\]
Substitute the above value in equation (2)
\[\tan y = \tan 2\theta \]
Cancel out tan on both sides from the above equation.
\[y = 2\theta \]
Put \[\theta = {\tan ^{ - 1}}t\]
\[y = 2{\tan ^{ - 1}}t\]
Differentiate y with respect to t.
We know that derivative of \[{\tan ^{ - 1}}x = \dfrac{1}{{(1 + {x^2})}}\]
Therefore, \[\dfrac{{dy}}{{dt}} = \dfrac{2}{{(1 + {t^2})}}\] --- equation (3)
Given: \[\sin \,x = \dfrac{{2\operatorname{t} }}{{\left( {1 + {t^2}} \right)}}\] --- equation (4)
Apply the same steps as mentioned above and find the value of \[\dfrac{{dx}}{{dt}}\] i.e. differentiating x with respect to t.
Now , Put \[t = \tan \theta \] in the given equation.
\[\operatorname{sinx} = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\] -- equation (5)
We know that \[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\] -- equation (6)
Comparing equation (5) and equation (6), we can write
\[\operatorname{sinx} = sin2\theta \]
Cancel out sin on both sides from the above equation.
Therefore, \[x = 2\theta \]
Substitute, \[\theta = {\tan ^{ - 1}}t\]
\[x = 2{\tan ^{ - 1}}t\]
Now differentiate x with respect to t.
\[\dfrac{{dx}}{{dt}} = \dfrac{2}{{(1 + {t^2})}}\]-- equation (7)
Divide equation (3) by equation (7) to get \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy/dt}}{{dx/dt}} = \dfrac{{\dfrac{2}{{(1 + {t^2})}}}}{{\dfrac{2}{{(1 + {t^2})}}}}\]
Simplify the above equation
\[\dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} = \dfrac{2}{{(1 + {t^2})}} \times \dfrac{{(1 + {t^2})}}{2}\]
Cancel out the like terms.
Therefore, \[\dfrac{{dy}}{{dx}} = 1\]

Hence, the correct option is c.

Note: There is an alternate method to solve the given problem. You can differentiate y with respect to t and x with respect to t without using any identities or formulas. While differentiating, make sure to apply the chain rule and the product or quotient rules correctly.