Prove the following

Solution:

Given,

sin x = 2t/(1 + t2), tan y = 2t/ (1 – t2)

Now, taking $\sin x=\frac{2 t}{1+t^{2}}$

Differentiating both sides w.r.t $t$, we get

$\cos x \cdot \frac{d x}{d t}=\frac{\left(1+t^{2}\right) \cdot \frac{d}{d t}(2 t)-2 t \cdot \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}$

$\cos x \cdot \frac{d x}{d t}=\frac{2\left(1+t^{2}\right)-2 t \cdot 2 t}{\left(1+t^{2}\right)^{2}}$

$\frac{d x}{d t}=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}} \times \frac{1}{\cos x}$

$\frac{d x}{d t}=\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}} \times \frac{1}{\sqrt{1-\sin ^{2} x}}$

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \times \frac{1}{\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}}$

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \times \frac{1}{\sqrt{\frac{\left(1+t^{2}\right)^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}}}$

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \times \frac{1+t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}}}$

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1 \quad^{2}\right)^{2}} \times \frac{\left(1+t^{2}\right)}{\sqrt{1+t^{4}-2 t^{2}}}$

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)} \times \frac{1}{\sqrt{\left(1-t^{2}\right)^{2}}}$

$\Rightarrow \quad \frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)} \times \frac{1}{\left(1-t^{2}\right)} \Rightarrow \frac{d x}{d t}=\frac{2}{1+t^{2}}$

Now taking, $\tan y=\frac{2}{1-t^{2}}$

Differentiating both sides w.r.t, $t$, we get

$\frac{d}{d t}(\tan y)=\frac{d}{d t}\left(\frac{2 t}{1-t^{2}}\right)$

$\sec ^{2} y \frac{d y}{d t}=\frac{\left(1-t^{2}\right) \cdot \frac{d}{d t}(2 t)-2 t \cdot \frac{d}{d t}\left(1-t^{2}\right)}{\left(1-t^{2}\right)^{2}}$

$\sec ^{2} y \frac{d y}{d t}=\frac{\left(1-t^{2}\right) \cdot 2-2 t \cdot(-2 t)}{\left(1-t^{2}\right)^{2}}$

$\sec ^{2} y \frac{d y}{d t}=\frac{2-2 t^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}}$

$\frac{d y}{d t}=\frac{2+2 t^{2}}{\left(1-t^{2}\right)^{2}} \times \frac{1}{\sec ^{2} y}$

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{1}{1+\tan ^{2} y}$

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{1}{1+\left(\frac{2 t}{1-t^{2}}\right)^{2}}$

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{1}{\frac{\left(1-t^{2}\right)^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}}}$

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{\left(1-t^{2}\right)^{2}}{1+t^{2}+2 t^{2}+4 t^{2}}$

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{\left(1-t^{2}\right)^{2}}{1+t^{4}+2 t^{2}}$

$\Rightarrow \frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{\left(1-t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}} \Rightarrow \frac{d y}{d t}=\frac{2}{1+t^{2}}$

$\therefore \frac{d y}{d t}=\frac{d y / d t}{d x / d t}=\frac{\frac{2}{1+t^{2}}}{\frac{2}{1+t^{2}}}=1$

Thus, $\frac{d y}{d t}=1$