Their Difference Is Just One!

$\frac{\tan^2 \theta}{\tan^2 \theta+1}=\frac{ \frac{\sin^2\theta}{\cos^2\theta} }{ \frac{\sin^2\theta}{\cos^2\theta} +1 }= \frac{ \frac{\sin^2\theta}{\cancel{\cos^2\theta}} }{\frac{\sin^2\theta+\cos^2\theta}{\cancel{\cos^2 \theta}}} = \sin^2\theta$

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$\sin^2\theta+\cos^2\theta=1$

In this problem we use the identities: $\tan \theta=\dfrac{\sin \theta}{\cos \theta}$ $\implies$ $\boxed{\tan^2 \theta=\dfrac{\sin^2 \theta}{\cos^2 \theta}}$ and $\boxed{\sin^2 \theta + \cos^2 \theta = 1}$

$\dfrac{\tan^2 \theta}{\tan^2 \theta + 1}=$ $\large \dfrac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\frac{\sin^2 \theta}{\cos^2\theta}+1}=\dfrac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}}=\dfrac{\sin^2 \theta}{\cos^2 \theta} \times \dfrac{\cos^2 \theta}{\sin^2 \theta+\cos^2 \theta}=\dfrac{\sin^2 \theta}{1}=$ $\color{#D61F06}\large \boxed{\sin^2 \theta}$

$\begin{aligned} \dfrac{\tan^2 \theta}{\tan^2 \theta + 1} &= 1 - \dfrac {1}{\sec^2 \theta}\\ &= 1- \cos^2 \theta\\ &= \sin^2 \theta \end{aligned}$

$\frac{\tan^2(\theta)}{\tan^2(\theta) + 1}$

$\frac{\tan^2(\theta)}{\frac{\sin^2(\theta)}{\cos^2(\theta)} + 1}$

$\frac{\tan^2(\theta)}{\frac{1 - \cos^2(\theta)}{\cos^2(\theta)} + 1}$

$\frac{\tan^2(\theta)}{\frac{1 - \cos^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)}}$

$\frac{\tan^2(\theta)}{\frac{1}{\cos^2(\theta)}}$

$\tan^2(\theta)\cos^2(\theta)$

$\frac{\sin^2(\theta)}{\cos^2(\theta)}\cos^2(\theta)$

$\large \boxed{\sin^2(\theta)}$

Since tan^2 θ +1 = sec^2 θ

so, tan^2 θ / (tan^2 θ +1) =tan^2 θ/sec^2 θ =sin^2 θ / cos^2θ /1/cos^2 θ = sin^2 θ