Let tan y-2t1-t2 and sin x-2t1-t2- Then find dydx-

Now, tan y=2t1 t^2 and sin x=2t1+t^2 or, y=tan^ 12t1 t^2 and x=sin^ 12t1+t^2 or, y=2tan^ 1 t and x=2tan^ 1 t or, y=x .This ...

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Integration by Parts

Let tan y=2...

Question

Let $tan y = \frac{2 t}{1 - t^{2}}$ and $sin x = \frac{2 t}{1 + t^{2}}$ . Then find $\frac{d y}{d x}$ .

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s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}},

\frac{d y}{d x}

| t | < 1

sin x = \frac{2 t}{1 + t^{2}}

tan y = \frac{2 t}{1 - t^{2}}

\frac{d y}{d x}

$s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}}$

Find $\frac{d y}{d x}$

sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}}

sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}}

\frac{d y}{d x}

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