MHT CET · Maths · Differentiation
If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is
- A 1
- B \(t\)
- C \(\frac{1}{1-t}\)
- D \(\frac{1}{1+t}\)
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
Given, \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\)
Now, \(x=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)\)
\(x=2 \tan ^{-1} t\)..(i)
and \(\quad y=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\)
\(y=2 \tan ^{-1} t\)...(ii)
From Eq. (i), \(\frac{d x}{d t}=\frac{2}{1+t^{2}}\)
From Eq. (ii), \(\frac{d y}{d t}=\frac{2}{1+t^{2}}\)
\(\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{2}{\left(1+t^{2}\right)} \times \frac{\left(1+t^{2}\right)}{2}=1\)
Now, \(x=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)\)
\(x=2 \tan ^{-1} t\)..(i)
and \(\quad y=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\)
\(y=2 \tan ^{-1} t\)...(ii)
From Eq. (i), \(\frac{d x}{d t}=\frac{2}{1+t^{2}}\)
From Eq. (ii), \(\frac{d y}{d t}=\frac{2}{1+t^{2}}\)
\(\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{2}{\left(1+t^{2}\right)} \times \frac{\left(1+t^{2}\right)}{2}=1\)
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