<s>[INST] If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$ 
If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$

If we take $t=\tan \theta$ then $x=2\theta$ and for $y$  ,I am getting two values as follows ,
if we apply identity $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$ then we get $y=2\theta$
and if we take $\tan \theta $ common from $\dfrac{1-\tan^2 \theta}{2\tan\theta}$  and simplify then we get $y= \dfrac{\pi}{4} + \theta$ ?
how we are getting two values of $y$ ?
thankyou
 [/INST] I'm not sure where you're getting $y=\theta+\pi/4$ but it is definitely incorrect as $$\cot(\theta+\pi/4)=\frac{1-\tan\theta\tan\pi/4}{\tan\theta+\tan\pi/4}=\frac{1-\tan\theta}{1+\tan\theta}$$ is not the same as $$\frac{1-t^2}{2t}=\frac{1-\tan^2\theta}{2\tan\theta}.$$
</s>